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v 


WORKS  OF  H.  B.  PHILLIPS, 

PH.D. 

PUBLISHED    BY 

JOHN  WILEY  &  SONS,   Inc. 

Differential  Equations. 

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DIFFERENTIAL  CALCULUS 


BY 


H.  B.  PHILLIPS,   Ph.D. 

Assistajit  Professor  of  Mathematics  in  the  Massachusetts 
Institute  of  Technology 


TOTAL  ISSUE    TEN  THOUSAND 


NEW  YORK 
JOHN    WILEY    &   SONS,    Inc. 

London:   CHAPMAN  &  HALL,  Limited 


•       •    < 

•  •      •    • 


•     •    -  «    1 


<      .       I  • 


•        •    •      • 

•   »      e  < 


Engineering 
Library 


Copyright,  1916, 

BY 

H.  B.  PHILLIPS 


Stanbopc  jptess 

F     H.GILSON   COMPANY 
BOSTON,  U.S.A. 


10-22 


PREFACE 


In  this  text  on  differential  calculus  I  have  continued  the 
plan  adopted  for  my  Analytic  Geometry,  wherein  a  few  cen- 
tral methods  are  expounded  and  applied  to  a  large  variety 
of  examples  to  the  end  that  the  student  may  learn  principles 
and  gain  power.  In  this  way  the  differential  calculus  makes 
only  a  brief  text  suitable  for  a  term's  work  and  leaves  for  the 
integral  calculus,  which  in  many  respects  is  far  more  impor- 
tant, a  greater  proportion  of  time  than  is  ordinarily  devoted 
to  it. 

.  As  material  for  review  and  to  provide  problems  for  which 
answers  are  not  given,  a  supplementary  list,  containing  about 
half  as  many  exercises  as  occur  in  the  text,  is  placed  at  the 
end  of  the  book. 

I  wish  to  acknowledge  my  indebtedness  to  Professor  H.  W. 
Tyler  and  Professor  E.  B.  Wilson  for  advice  and  criticism 
and  to  Dr.  Joseph  Lipka  for  valuable  assistance  in  preparing 
the  manuscript  and  revising  the  proof. 

H.  B.  PHILLIPS. 

Boston,  Mass.,  August,  1916. 


2  1545 

•  •• 

in 


CONTENTS 


Chapter  Pages 

1.   Introduction 1_    9 

I.   Derivative  and  Differential 10-  18 

1- _—  HI,   Differentiation  of  Algebraic  Functions 19-  31 

IV.   Rates 32-38 

V.   Maxima  and  Minima 39-  48 

VI.   Differentiation  of  Transcendental  Functions.  49-  62 

VII.   Geometrical  Applications 63_  84 

>U-  VIII.   Velocity  and  Acceleration  in  a  Curved  Path  .  85-  93 

IX.   Rolle's  Theorem  and  Indeterminate  Forms 94-100 

X.   Series  and  Approximations 101-112 

XI.    Partial  Differentiation 113-139 

Supplementary  Exercises 140-153 

Answers 154-160 

Index 161"162 


• 


• 


DIFFERENTIAL  CALCULUS 


CHAPTER  I 
INTRODUCTION 

1.  Definition  of  Function.  —  A  quantity  y  is  called  a 
Junction  of  a  quantity  x  if  values  of  y  are  determined  by  values 
of  x. 

Thus,  if  y  =  1  —  x2,  y  is  a  function  of  x;  for  a  value  of  x 
determines  a  value  of  y.  Similarly,  the  area  of  a  circle  is  a 
function  of  its  radius;  for,  the  radius  being  given,  the  area  is 
determined. 

It  is  not  necessary  that  only  one  value  of  the  function 
correspond  to  a  value  of  the  variable.  Several  values  may  be 
determined.     Thus,  if  x  and  y  satisfy  the  equation 

x2  —  2  xy  +  y2  =  x, 

then  y  is  a  function  of  x.  To  each  value  of  x  correspond  two 
values  of  y  found  by  solving  the  equation  for  y. 

A  quantity  u  is  called  a  function  of  several  variables  if  u  is 
determined  when  values  are  assigned  to  those  variables. 

Thus,  if  z  =  x2  +  y2,  then  z  is  a  function  of  x  and  y;  for, 
values  being  given  to  x  and  y,  a  value  of  z  is  determined. 
Similarly,  the  volume  of  a  cone  is  a  function  of  its  altitude 
and  radius  of  base;  for  the  radius  and  altitude  being  assigned, 
the  volume  is  determined. 

2.  Kinds  of  Functions.  —  An  exjDressipn  containing 
vjmables  is-caliejj.  &JL  explicit  function  qf_  those  variables. 
Thus  Vx  +  y  is  an  explicit  function  of  x  and  y.     Similarly,  if 

y  =  Vx  +  1, 
y  is  an  explicit  function  of  x. 


2  DIFFERENTIAL  CALCULUS  Chap.  I. 

(        «     r         ( 

(        (  i 

1  f       '     <         ,  <  i        , 

r  ,  '      '         .      '      ,  ' 

A  quantity  determined  by  an  equation  not  solved  for  that 
quantity  is  called  an  implicit  function.     Thus,  if 

x2  —  2  xy  +  y2  =  x, 

y  is  an  implicit  function  of  x.    Also  x  is  an  implicit  function 
of  y. 

Explicit  and  implicit  do  not  denote  properties  of  the  func- 
tion but  of  the  way  it  is  expressed.  An  implicit  function  is 
rendered  explicit  by  solving.  For  example,  the  above  equa- 
tion is  equivalent  to 

y  =  x  d=  Va;, 
in  which  y  appears  as  an  explicit  function  of  x. 

A  rational  function  is  one  representable  by  an  algebraic 
expression  containing  no  fractional  powers  of  variable  quanti- 
ties.    For  example, 

xVE  +  3 
x2  +  2x 

is  a  rational  function  of  x. 

An  irrational  function  is  one  represented  by  an  algebraic 
expression  which  cannot  be  reduced  to  rational  form.     Thus 

Va;  +  y  is  an  irrational  function  of  x  and  y. 

A  function  is  called  algebraic  if  it  can  be  represented  by  an 
algebraic  expression  or  is  the  solution  of  an  algebraic  equa- 
tion.    All  the  functions  previously  mentioned  are  algebraic. 

Functions  that  are  not  algebraic  are  called  transcendental. 
For  example,  sin  x  and  log  x  are  transcendental  functions  of  x. 

3.  Independent  and  Dependent  Variables.  —  In  most 
problems  there  occur  a  number  of  variable  quantities  con- 
nected by  equations.  Arbitrary  values  can  be  assigned  to 
some  of  these  quantities  and  the  others  are  then  determined. 
Those  taking  arbitrary  values  are  called  independent  vari- 
ables; those  determined  are  called  dependent  variables. 
Which  variables  are  taken  as  independent  and  which  as  de- 
pendent is  usually  a  matter  of  convenience.  The  number  of 
independent  variables  is,  however,  determined  by  the  equa- 
tions. 


Chap.  I.  INTRODUCTION  3 

For  example,  in  plotting  the  curve 

y  =  xz  +  x, 

values  are  assigned  to  x  and  values  of  y  are  calculated.  The 
independent  variable  is  x  and  the  dependent  variable  y.  We 
might  assign  values  to  y  and  calculate  values  of  x  but  that 
would  be  much  more  difficult. 

4.  Notation.  —  A  particular  function  of  x  is  often  repre- 
sented by  the  notation  /  (x),  which  should  be  read,  function 
of  x,  or  /  of  .r,  not  /  times  x.     For  example, 

/  (x)  =  Vx2  +  1 

means  that/  {x)  is  a  symbol  for  Vz2  +  1.     Similarly, 

V  =  f  (x) 

means  that  y  is  some  definite  (though  perhaps  unknown) 
function  of  x. 

If  it  is  necessary  to  consider  several  functions  in  the  same 
discussion,  they  are  distinguished  by  subscripts  or  accents  or 
by  the  use  of  different  letters.  Thus,  /i  (x),  /2  (x),  f  (x), 
f"  (x),  g  (x)  (read  /-one  of  x,  /-two  of  x,  /-prime  of  x,  /-second 
of  .x,  g  of  x)  represent  (presumably)  different  functions  of  x. 

Functions  of  several  variables  are  expressed  by  writing 
commas  between  the  variables.     For  example, 

v  =  f  (r,  h) 

expresses  that  v  is  .a  function  of  r  and  h  and 

v  =  f  (a,  6,  c) 

expresses  that  v  is  a  function  of  a,  6,  c 

The  /  in  the  symbol  of  a  function  should  be  considered  as 
representing  an  operation  to  be  performed  on  the  variable  or 
variables.     Thus,  if 

/  (x)  =  Vx2  +  1, 

/represents  the  operation  of  squaring  the  variable,,  adding  1, 
and  extracting  the  square  root  of  the  result.     If  x  is  replaced 


4  DIFFERENTIAL  CALCULUS  Chap.  I. 

by  any  other  quantity,  the  same  operation  is  to  be  performed 
on  that  quantity.     For  example, 

/  (2)  =  V22  +  1  =  V5. 
f{y+  1)  =  ^(2/+ 1)2+1  =  Vtf  +  2y  +  2. 
Similarly,  if 

/  (x,  y)  =  x2  +  xy  -  y2, 

then  /  (1,  2)  =  l2  +  1  •  2  -  22  =  -1. 

If  /  (x,  y,  z)  =  x2  +  y2  +  z2, 

then  f  (2,  -3,  1)  =  22  +  (-3)2  +  1  =  14. 

EXERCISES 


333 

2    1      2         2 


1.  Given  x   +  y    =  a  ,  express  2/  as  ari  explicit  function  of  #. 

2.  Given  logio  (x)  =  sin  ?/,  express  x  as  an  explicit  function  of  y. 
Also  express  ?/  as  an  explicit  function  of  x. 

3.  If  /  (x)  =  x2  -  3  x  +  2,  show  that  /  (1)  =  /  (2)  =  0. 

4.  If  F  (x)  =  x*  +  2  x2  +  3,  show  that  F  (-a)  *=,  i*1  (a). 

5.  If  F  (re)  =  x  +  -,  find  F(x  +  1).     AlsQ«nnci  F  (#  +  1. 

6.  If  <f>  (x)  =  Vx2  -  1,  find  4>{2x).     Also  find  2  <£  (.c). 

7.  If  ^  (x)  =  0>/C  +  20,  find  I  f  -Y     Also  find  -)-• 

8.  If  A  (jb)  =  2*,/2  (3)  =  x2,  find/!  [/2  (y)].     Also  find/2  [A  (y)]. 

9.  If/  (x,  y)  =  x  -  -,  show  that/  (2,  1)  =  2/(1,  2)  =  1. 

10.  Given  /  (x,  y)  =  x2  +  .r?/,  find  /  (y,  x) . 

11.  On  how  many  independent  variables  docs  the  volume  of  a  right 
circular  cylinder  depend?  - 

12.  Three  numbers  x,  y,  z  satisfy  two  equations 

x2  -\-  y2  -{-  z2  =  5, 
x  +  y   +Z    =  1. 

How  many  of  these  numbers  can  be  taken  as  independent  variables? 

5.  Limit.  —  If  in  any  process  a  variable  quantity  ap- 
proaches a  constant  one  in  such  a  way  that  the  difference  of 
the  two  becomes  and  remains  as  small  as  you  please,  the  con- 
stant is  said  to  be  the  limit  of  the  variable. 

The  use  of  limits  is  well  illustrated  by  the  incommensurable 


Chap.  I.  INTRODUCTION  5 

cases  of  geometry  and  the  determination  of  the  area  of  a 
circle  or  the  volume  of  a  cone  or  sphere. 

6.  Limit  of  a  Function.  —  As  a  variable  approaches  a 
limit  a  function  of  that  variable  may  approach  a  limit.  Thus, 
as  x  approaches  1,  x2  +  1  approacnes  2. 

We  shall  express  that  a  variable  x  approaches  a  limit  a  by 

the  notation 

x  =  a. 

The  symbol  =  thus  means  "approaches  as  a  limit." 

Let  /  (x)  approach  the  limit  A  as  x  approaches  a;  this  is 

expressed  by 

lim/(:r)  =  A, 

x=a 

which  should  be  read,  "  the  limit  of  /  (x),  as  x  approaches  a, 
is  .4." 

Example  1.   Find  the  value  of 

lim 


lim  ( x  H — )• 
x=l  V  xl 


As  x  approaches  1,  the  quantity  x  +  -  approaches  1  +  - 


or  2.     Hence 


lim  (x  +  -)  =  2. 

x=l\  x) 


Ex.  2.   Find  the  value  of 

,.  sin  6 

lim 


[6=0    1  +  COS  0 

As  6  approaches  zero,  the  function  given  approaches 

f0 


1  +  1 
Hence 


=  0. 


,.  sin  0  rt 

hm  —- =  0. 

0=0  1  +  cos  0 

7.  Properties  of  Limits.  —  In  finding  the  limits  of  func- 
tions frequent  use  is  made  of  certain  simple  properties  that 
follow  almost  immediately  from  the  definition. 


6  DIFFERENTIAL  CALCULUS  Chap.  I. 

1.  The  limit  of  the  sum  of  a  finite  number  of  functions  is 
equal  to  the  sum  of  their  limits. 

Suppose,  for  example,  X,  F,  Z  are  three  functions  ap- 
proaching the  limits  A,  B,  C  respectively.  Then  X+F+Z 
is  approaching  A  +  B  -J-  C.     Consequently, 

Km(X+Y  +  Z).=  A+B  +  C  =  lim  X  +  lim  Y  +  lim  Z. 

2.  The  limit  of  the  product  of  a  finite  number  of  functions 
is  equal  to  the  product  of  their  limits. 

If,  for  example,  X,  F,  Z  approach  A,  B,C  respectively, 
then  XYZ  approaches  ABC,  that  is, 

lim  XYZ  =  ABC  =  lim  X  lim  Y  lim  Z. 

3.  //  the  limit  of  the  denominator  is  not  zero,  the  limit  of  the 
ratio  of  two  functions  is  equal  to  the  ratio  of  their  limits. 

Let  X,  Y  approach  the  limits  A,  B  and  suppose  B  is  not 

X  A 

zero.     Then  ^  approaches  75 ,  that  is, 

X      A       limX 
hmY~  B      limF' 

If  B  is  zero  and  A  is  not  zero,  ^  will  be  infinite.     Then 

X  A  X 

^  cannot  approach  ^  as  a  limit;  for,  however  large  ^  may 

X 

become,  the  difference  of  ^and  infinity  will  not  become  small. 

8.    The  Form  -.  — When  x  is  replaced  by  a  particular 

value,  a  function  sometimes  takes  the  form  -•     Although  this 

symbol  does  nob  represeut  a  definite  value,  the  function  may 
have  a  definite  limit.     This  is  usually  made  evident  by  writ- 
ing the  function  in  a  different  form. 
Example  1.   Find  the  value  of 

,.     x2  —  1 

lim - 

x=l   x  —  1 


Chap.  I.  INTRODUCTION  7 

When  x  is  replaced  by  1,  the  function  takes  the  form 

1-1^0 

1-  1  ~0" 

Since,  however, 

x*-l 

7   =  X  +  1, 

X  —  \ 

the  function  approaches  1  -f  1  or  2.     Therefore 

x2  —  1 
lim- f  =  2. 

x  =  l   X  -  1 

Ex.  2.   Find  the  value  of 

..     (Vl  +  x  -  1) 
Inn 

2  =  0  x 

When  x  =  0  the  given  function  becomes 

1-1^0 

o    ~o' 

Multiplying  numerator  and  denominator  by  Vl  +  x  -f  1, 
Vl  +  x-  1  =  a;  1 

^  x  (Vi  +  z  + 1)     Vi  +  z  +  l ' 

As  x  approaches  0,  the  last  expression  approaches  J.     Hence 

..     (Vl  +  x  -  1)      1 
hm L  =  -  • 

x=0  X  Z 

9.   Infinitesimal.  —  A   variable   approaching   zero  as   a 
limit  is  called  an  infinitesimal. 

Let  a  and  /3  be  two  infinitesimals.     If 

hm- 

is  finite  and  not  zero,  a  and  /3  are  said  to  be  infinitesimals  of 
the  same  order.     If  the  limit  is  zero,  a  is  of  higher  order  than 

/3.     If  the  ratio  -  approaches  infinity,  |8  is  of  higher  order 

than  a.     Roughly  speaking,  the  higher  the  order,  the  smaller 
the  infinitesimal. 


8  DIFFERENTIAL  CALCULUS  Chap.  I. 

For  example,  let  x  approach  zero.     The  quantities 

/y»        /-y»2        /y»3        /y»4        r±4-  r% 

are  infinitesimals  arranged  in  ascending  order.     Thus  xA  is  of 
higher  order  than  x2;  for 

rr4 
lim  —  =  lim  x2  =  0. 
x±oxz      x=o 

Similarly,  x3  is  of  lower  order  than  x*,  since 

XA        X 

approaches  infinity  when  x  approaches  zero. 

As  x  approaches  ~ ,  cos  x  and  cot  x  are  infinitesimals  of  the 
same  order;  for 

lim  — i —  =  lim  sin  a;  =  1, 
>7rcotz      *=o 

which  is  finite  and  not  zero. 

EXERCISES 

Find  the  values  of  the  following  limits: 

„     v     rc2-2x  +  3  ...     Vl  -  a;2  -  Vl  +  x2 

1.  hm r 4.   lim - 

x=0        x  —  5  x=o  % 

n    ,.         sin  0  +  cos  0 

2.  lim  - — x— — : ^-z-  ..      sin  0 

v  sin  2  0  +  cos  2  0  5.   lim • 

e=2  0=0  tan  0 

3.  ft,^ -■'«  +  »■  •fi.  Iim-2S»  . 

x=i       a;  —  1  0=0  sin  2  0 

7.  By  the  use  of  a  table  of  natural  sines  find  the  value  of 

. .       sin  x 
lim   • 

a:=0       Z 

8.  Define  as  a  limit  the  area  within  a  closed  curve. 

9.  Define  as  a  limit  the  volume  within  a  closed  surface. 

10.  Define  V2. 

11.  On  the  segment  PQ  (Fig.  9a)  construct  a  series  of  equilateral  tri- 
angles reaching  from  P  to  Q.    As  the  number  of  triangles  is  increased, 


Chap.  I. 


INTRODUCTION 


their  bases  approaching  zero,  the  polygonal  line  PABC,  etc.,  approaches 
PQ.     Does  its  length  approach  that  of  PQ? 


A       C 


-\        A        A         A         A 

/\/\/N/\'*/% 

»/        \/        \>        \,'       \  /        \/      \ 


Q 


Fig.  9a. 


12.  Inscribe  a  series  of  cylinders  in  a  cone 
as  shown  in  Fig.  9b.  As  the  number  of  cyl- 
inders increases  indefinitely,  their  altitudes 
approaching  zero,  does  the  sum  of  the  vol- 
umes of  the  cylinders  approach  that  of  the 
cone?  Does  the  sum  of  the  lateral  areas  of 
the  cylinders  approach  the  lateral  area  of  the 
cone? 


Fig.  9b. 


13.  Show  that  when  x  approaches  zero,  tan  -  does  not  approach  a 

limit.  

14.  As  x  approaches  1,  which  of  the  infinitesimals  1  —  x  and  Vl  —  x 
is  of  higher  order? 

15.  As  the  radius  of  a  sphere  approaches  zero,  show  that  its  volume 
is  an  infinitesimal  of  higher  order  than  the  area  of  its  surface  and  of  the 
same  order  as  the  volume  of  the  circumscribing  cylinder. 


CHAPTER  II 


DERIVATIVE   AND   DIFFERENTIAL 

10.  Increment.  —  When  a  variable  changes  value,  the 
algebraic  increase  (new  value  minus  old)  is  called  its  in- 
crement and  is  represented  by  the  symbol  A  written  before 
the  variable. 

Thus,  if  x  changes  from  2  to  4,  its  increment  is 

Ax  =  4  -  2  =  2. 


If  x  changes  from  2  to  —1, 

Ax  =  -1 


2=  -3. 


When  the  increment  is  positive  there  is  an  increase  in 
value,  when  negative  a  decrease. 

Let  y  be  a  function  of  x.     When  x  receives  an  increment 

Ax,  an  increment  Ay  will  be 
determined.  The  increments 
of  x  and  y  thus  correspond. 
To  illustrate  this  graphically 
let  x  and  y  be  the  rectangular 
coordinates  of  a  point  P.  An 
equation 

y  =f  0) 

represents  a  curve.     When  x 

changes,  the  point  P  changes 

to  some  other  position  Q  on  the  curve.     The  increments  of 

x  and  y  are 

Ax  =  PR,         Ay  =  RQ.  (10) 

11.  Continuous  Function.  —  A  function  is  called  con- 
tinuous if  the  increment  of  the  function  approaches  zero  as 
the  increment  of  the  variable  approaches  zero. 

10 


Fig.  10. 


Chap.  II.         DERIVATIVE   AND   DIFFERENTIAL 


11 


In  Fig.  10,  y  is  a  continuous  function  of  x;  for,  as  Aa; 
approaches  zero,  Q  approaches  P  and  so  Ay  approaches  zero. 

In  Figs.  11a  and  lib  are  shown  two  ways  that  a  function 
can  be  discontinuous.     In  Fig.  11a  the  curve  has  a  break  at 


o 


Fig.  11a. 


Fig.  lib. 


P.  As  Q  approaches  P',  Ax  =  PR  approaches  zero,  but 
A?/  =  RQ  does  not.  In  Fig.  lib  the  ordinate  at  x  =  a  is 
infinite.  The  increment  Ay  occurring  in  the  change  from 
x  =  a  to  any  neighboring  value  is  infinite. 

12.  Slope  of  a  Curve.  —  As  Q  moves  along  a  continuous 
curve  toward  P,  the  line  PQ 
turns  about  P  and  usually 
approaches  a  limiting  posi- 
tion PT.  This  line  PT  is 
called  the  tangent  to  the 
curve  at  P. 

The  slope  of  PQ  is 

RQ  =  Ay 

PR      Ax' 

As  Q  approaches  P,  Ax  ap- 
proaches zei»o  and  the  slope 
of  PQ  approaches  that  of  PT.     Therefore 


Fig.  12a. 


Slope  of  the  tangent  =  tan  </>  =  lim  -^ 

Ax=0  Az 


(12) 


12 


DIFFERENTIAL  CALCULUS 


Chap.  II. 


The  slope  of  the  tangent  at  P  is  called  the  slope  of  the  curve 
at  P. 

Example.  Find  the 
slope  of  the  parabola 
y  =  x2  at  the  point  (1,  1). 

Let  the  coordinates  of 
P  be  x,  y.  Those  of 
Q  are  x  +  Ax,  y  +  Ay. 
Since  P  and  Q  are  both 
on  the  curve, 


y  =  x2 


and 


y  +  Ay  = (x  +  Ax)2  = 
x2  +  2  x  Ax  +  (Ax)2. 


Subtracting  these  equations,  we  get 

A?/  =  2  x  Ax  +  (Ax)2. 

Dividing  by  Ax, 

Ay 


Ax 


=  2  x  +  Ax. 


As  Ax  approaches  zero,  this  approaches 

Slope  at  P  =  2  x. 

This  is  the  slope  at  the  point  with  abscissa  x.     The  slope  at 
(1,  1)  is  then  2-1=2. 


13.   Derivative.  —  Let   y   be   a   function   of   x.     If 


A]/ 
Ax 


approaches  a  limit  as  Ax  approaches  zero,  that  limit  is  called 

the  derivative  of  y  with  respect  to  x.     It  is  represented  by  the 

notation  Dxy,  that  is, 

Ay 


Dxy  =  lim  -r— 
XJ      Ax±o  Ax 


(13a) 


If  a  function  is  represented  by  /(x),  its  derivative  with 
respect  to  x  is  often  represented  by  /'  (x).     Thus 


/'  (*)  =  lim  ^-  =  DJ  (x). 

Ax=0       ^X 


(13b) 


Chap.  II. 


DERIVATIVE   AND   DIFFERENTIAL 


13 


In  Art.  12  we  found  that  this  limit  represents  the  slope  of 
the  curve  y  =  f(x).  The  derivative  is,  in  fact,  a  function 
of  x  whose  value  is  the  slope  of 
the  curve  at  the  point  with  ab- 
scissa x. 

The  derivative,  being  the  limit 

of  .  - ,   is  approximately  equal 

AX 

to  a  small  change  in  y  divided 
by  the  corresponding  small 
change  in  x.  It  is  then  large  or 
small  according  as  the  small  in- 
crement of  y  is  large  or  small  in 
comparison  with  that  of  x. 
If  small  increments  of  x  and 


Ay 


Fig.  13. 


y  have   the  same   sign  -r—  and 

its  limit  Dxy  are  positive.  If  they  have  opposite  signs  Dxy 
is  negative.  Therefore  Dxy  is  positive  when  x  and  y  increase 
and  decrease  together  and  negative  when  one  increases  as  the 
other  decrease^ 

Example,     y  =  x?  —  3  x  +  2. 

Let  x  receive  an  increment  Ax.  The  new  value  of  x  is 
x  +  Ax.  The  new  value  of  y  is  y  -f-  Ay.  Since  these  satisfy 
the  equation, 

y  +  Ay  =  (x  +  Ax)3  -  3  (x  -f  Ax)  +  2. 
Subtracting  the  equation 

y  =  x?-3x+2, 
we  get 

Ay  =  3  x2  Ax  +  3  x  (Ax)2  +  (Az)3  -  3  Ax. 

Dividing  by  Ax, 

^=3x2  +  3xAx-f-  (Ax)2  -  3. 

As  Ax  approaches  zero  this  approaches  the  limit 

Dxy  =  3  x2  -  3. 


14  DIFFERENTIAL  CALCULUS  Chap.  II. 

The  graph  is  shown  in  Fig.  13.  At  A  (where  x  —  1)  y  =  0 
and  Dxy  =  3  •  1  —  3  =  0.  The  curve  is  thus  tangent  to  the 
#-axis  at  A.  The  slope  is  also  zero  at  B  (where  x  =  —  1). 
This  is  the  highest  point  on  the  arc  AC.  On  the  right  of  A 
and  on  the  left  of  B,  the  slope  Dxy  is  positive  and  x  and  y  in- 
crease and  decrease  together.  Between  A  and  B  the  slope 
is  negative  and  y  decreases  as  x  increases. 

<£- — ^  EXERCISES 

v£  Given  y  =  v  x,  find  the  increment  of  y  when  x  changes  from 
x  =  2  to  x  =  1.9.  Show  that  the  increments  approximately  satisfy 
the  equation 

Ay  =     1_ 

Az  "  2  Vx 
•  2.   Given  y  =  logio  x,  find  the  increments  of  y  when  x  changes  from 
50  to  51  and  from  100  to  101.     Show  that  the  second  increment  is  ap- 
proximately half  the  first. 

3.  The  equation  of  a  certain  line  is  y  =  2  x  -f-  3.     Find  its  slope  by 

calculating  the  limit  of  —  • 

Ax 

4.  Construct  the  parabola  y  =  x2  —  2  x.  Show  that  its  slope  at 
the  point  with  abscissa  x  is  2  (x  —  1).  Find  its  slope  at  (4,  8).  At 
what  point  is  the  slope  equal  to  2? 

V5.  Construct  the  curve  represented  by  the  equation  y  =  x*  —  2  x2. 
Show  that  its  slope  at  the  point  with  abscissa  x  is  4  x  (x2  —  1).  At  what 
points  are  the  tangents  parallel  to  the  x-axis?  Indicate  where  the  slope 
is  positive  and  where  negative. 

In  each  of  the  following  exercises  show  that  the  derivative  has  the 
value  given.     Also  find  the  slope  of  the  corresponding  curve  at  x  =  —  1. 

6.  y  =  (x  +  1)  (x  +  2),     Dxy  =  2  x  +  3. 

7.  y  =  x\  Dxy  =  4  xz. 

8.  y  =  x3  -  x2,  Dxy  =  3  x2  -  2  x. 

9.  y  =  -,  Dxy  =  --• 

10.  If  x  is  an  acute  angle,  is  Dx  cos  x  positive  or  negative? 

11.  For  what  angles  is  Dx  sin  x  positive  and  for  what  angles  negative? 

14.  Approximate  Value  of  the  Increment  of  a  Function. — 
Let  y  be  a  function  of  x  and  represent  by  e  a  quantity  such 

that  At/        n       . 


Chap.  II.         DERIVATIVE  AND   DEFERENTIAL  15 

As  Ax  approaches  zero,  -r^  approaches  Dxy  and  so  e  ap- 

fcJifr 

proaches  zero. 

The  increment  of  y  is 

Ay  =  Dxy  Ax  +  eAx. 
The  part 

Dxy  Ax  (14) 

is  called  the  principal  part  of  Ay.  It  differs  from  Ay  by  an 
amount  eAx.  As  Ax  approaches  zero,  e  approaches  zero,  and 
so  eAx  becomes  an  indefinitely  small  fraction  of  Ax.  It  is  an 
infinitesimal  of  higher  order  than  Ax.  If  then  the  principal 
part  is  used  as  an  approximation  for  Ay,  the  error  will  be 
only  a  small  fraction  of  Ax  when  Ax  is  sufficiently  small. 

Example.  When  x  changes  from  2  to  2.1  find  an  approxi- 
mate value  for  the  change  in  y  =  -• 

x 

In  exercise  9,  page  14,  the  derivative  of  -  was  found  to 

x 

be -2.     Hence  the  principal  part  of  Ay  is 

Jb 

-\Ax=  -j(-1)  -  -0.0250. 
x2  4 

The  exact  increment  is 

Ay  =  g§F  ~  I  =  -°-0232- 

The  principal  part  represents  Ay  with  an  error  less  than  0  002 
which  is  2%  of  Ax. 

10.  Differentials.  —  Let  x  be  the  independent  variable 
and  let  y  be  a  function  of  x.  The  principal  part  of  Ay  is 
called  the  differential  of  y  and  is  denoted  by  dy;  that  is, 

dy  =  Dxy  Ax.  (15a) 

This  equation  defines  the  differential  of  any  function  y  of  x. 
In  particular,  if  y  =  x,  Dxy  =  1,  and  so 

dx  =  Ax,  (15b) 

that  is,  the  differential  of  the  independent  variable  is  equal  to 


16 


DIFFERENTIAL  CALCULUS 


Chap.  II. 


its  increment  and  the  differential  of  any  function  y  is  equal  to 
the  product  of  its  derivative  and  the  increment  of  the  independent 
variable. 

Combining  15a  and  15b,  we  get 


whence 


dy  =  Dxy  dx, 

t  -  ** 


(15c) 
(15d) 


dy 


that  is,  the  quotient  ~  is  equal  to  the  derivative  of  y  with  respect 

to  x. 

Since  Dxy  is  the  slope  of  the  curve  y  =  f  (x),  equations  15b 

and  15c  express  that  dy  and  dx  are  the  sides  of  the  right  tri- 
angle PRT  (Fig.  15)  with 
hypotenuse  PT  extending 
along  the  tangent  at  P. 
On  this  diagram,  Ax  and  Ay 
are  the  increments 


Ax  =  PR, 

occurring    in 
from  P  to  Q. 
tials  are 


Ay  =  RQ, 

the    change 

The  differen- 


ce =  PR,        dy  =  RT. 

A  point  describing  the 
curve  is  moving  when  it 
passes  through  P  in  the  direction  of  the  tangent  PT.  The 
differential  dy  is  then  the  amount  y  would  increase  when  x 
changes  to  x  +  Ax  if  the  direction  of  motion  did  not  change. 
In  general  the  direction  of  motion  does  change  and  so  the 
actual  increase  Ay  =  RQ  is  different  from  dy.  If  the  in- 
crements are  small  the  change  in  direction  will  be  small  and 
so  Ay  and  dy  will  be  approximately  equal. 

Equation  15c  was  obtained  under  the  assumption  that  x 
was  the  independent  variable.  It  is  still  valid  if  x  and  y  are 
continuous  functions  of  an  independent  variable  t.     For  then 

dx  =  Dtx  At,        dy  =  Dty  At. 


Chap.  II.         DERIVATIVE  AND   DIFFERENTIAL 

The  identity 

A//      Ay     Ax 

At    ~  Ax*  At 

r 

gives  in  the  limit 

Dty  =  Dxy  •  Dtx. 

Hence 

that  is, 

Dty  At  =  Dxy  -  Dtx  At, 

dy  =  Dxy  dx. 

Example  1. 

Given  y  =  — —  ,  find  dy. 

X 

In  this  case 

Ay  = 

a;  +  Arc  +  1      x  +  1                 Ax 

x  +  Ax             x              x  (x  +  Ax) 

Consequently, 

Ay                  1 

Ax          x  (x  +  Ax) 

17 


As  Arc  approaches  zero,  this  approaches 

dy  =  _  1^ 
dx  a;2' 

Therefore 

#r.  2.   Given  x  =  t\  y  =  t\  find  ^  • 

The  differentials  of  x  and  ?/  are  found  to  be 

dx  =  2t  dt,        dy  =  3  t2  dt. 

Division  then  gives, 

dy      3 

dx"  2 

2£c.  3.  An  error  of  1%  is  made  in  measuring  the  side  of  a 
square.     Find  approximately  the  error  in  the  calculated  area. 

Let  x  be  the  correct  measure  of  the  side  and  x  +  Ax  the 
value  found   by  measurement.     Then  dx  =  Ax  =  =b0  01  x. 


nxn— l. 


J 
18  DIFFERENTIAL  CALCULUS  Chap.  II. 

The  error  in  the  area  is  approximately 

dA  =  d  (x2)  =2xdx=  ±0.02  x2  =  ±0.02  A, 

which  is  2%  of  the  area. 

EXERCISES 

1.  Let  n  be  a  positive  integer  and  y  =  xn.     Expand 

Ay  =  {x  +  Ax)n  -  xn 

by  using  the  binomial  theorem.     Show  that 

dy 
dx 

What  is  the  principal  part  of  Ay? 

2.  Using  the  results  of  Ex.  1,  find  an  approximate  value  for  the  in- 
crement of  x6  when  x  changes  from  1.1  to  1.2.     Express  the  error  as  a 

percentage  of  Ax. 

dA 

3.  If  A  is  the  area  of  a  circle  of  radius  r,  show  that  -j-  is  equal  to  the 

circumference. 

4.  If  the  radius  of  a  circle  is  measured  and  its  area  calculated  by 
using  the  result,  show  that  an  error  of  1%  in  the  measurement  of  the 

radius  will  lead  to  an  error  of  about  2%  in  the  area. 

dv 
6.    If  v  is  the  volume  of  a  sphere  with  radius  r,  show  that  -=-  is  equal 

to  the  area  of  its  surface. 

6.  Let  v  be  the  volume  of  a  cylinder  with  radius  r  and  altitude  h. 

dv 
Show  that  if  r  is  constant  -77  is  equal  to  the  area  of  the  base  of  cylinder 

dv 
and  if  h  is  constant  -7-  is  equal  to  the  lateral  area. 

7.  If  y  =  f  (x)  and  for  all  variations  in  x,  dx  =  Ax,  dy  =  Ay,  show 
that  the  graph  of  y  =  f  (x)  is  a  straight  line. 

8.  If  y  is  the  independent  variable  and  x  =  f  (y),  make  a  diagram 
showing  dx,  dy,  Ax,  and  Ay. 

9.  If  the  y-axis  is  vertical,  the  rr-axis  horizontal,  a  body  thrown  hori- 
zontally from  the  origin  with  a  velocity  of  50  ft.  per  second  will  in  t 
seconds  reach  the  point 

x  =  50*,        y  =  -16  t2. 

Find  the  slope  of  its  path  at  that  point. 

10.   A  line  turning  about  a  fixed  point  P  intersects  the  x-axis  at  A 

and  the  y-axis  at  B.     If  K\  and  K2  are  the  areas  of  the  triangles  OP  A  and 

OPB,  show  that 

dKi      PA* 

dK2  _  PB* 


CHAPTER  III 
DIFFERENTIATION    OF   ALGEBRAIC    FUNCTIONS 

16.  The  process  of  finding  derivatives  and  differentials 
is  called  differentiation.  Instead  of  applying  the  direct 
method  of  the  last  chapter,  differentiation  is  usually  per- 
formed by  means  of  certain  formulas  derived  by  that  method. 
In  this  work  we  use  the  letter  d  for  the  operation  of  taking  the 

differential  and  the  symbol  -r-  for  the  operation  of   taking 

the  derivative  with  respect  to  x.     Thus 
d  (u  +  v)  =  differential  of  (u  +  v), 

-3-  (u  +  v)  =  derivative  of  (u  +  v)  with  respect  to  x. 
ax 

To  obtain  the  derivative  with  respect  to  x  we  proceed  as  in 
finding  the  differential  except  that  d  is  everywhere  replaced 

1      d 

17.  Formulas.  —  Let  u,  v,  w  be  continuous  functions  of  a 
single  variable  x,  and  c,  n  constants.* 


I. 

dc  =  0. 

II. 

d  (u  -\-  v)  =  du  +  dv. 

III. 

d  (cu)  =  c  du. 

IV. 

d  (w)  =  u  dv  +  v  du. 

V. 

m  fu\        /*  du     ■  u  dv 

\v)                  vz 

VI.  d  (a")  =  nw"-1  du. 


*  It  is  assumed  that  the  functions  u,  v,  w  have  derivatives.     There 

exist  continuous  functions, 

u  =f(x), 

19 


20  DIFFERENTIAL  CALCULUS  Chap.  III. 

18.  Proof  of  I.  —  The  differential  of  a  constant  is  zero. 

When  a  variable  x  takes  an  increment  Ax,  a  constant  does 

Ac 
not  vary.     Consequently,  Ac  =  0,  -r-1  r=  0,  and  in  the  limit 

llx 

dc  ^- " 

—  =  o.     Clearing  of  fractions, 

dx 

dc  =  dx  •  0  =  0. 

19.  Proof  of  II.  —  The  differential  of  the  sum  of  a  finite 
number  of  functions  is  equal  to  the  sum  of  their  differentials. 

Let 

y  =  u  +  v. 

When  x  takes  an  increment  Ax,  u  will  change  to  u  +  Ait,  v 
to  v  +  Ay,  and  y  to  y  +  Ay.     Consequently 

y  +  Ay  =  u  +  Ait  +  v  +  Ay. 

Subtraction  of  the  two  equations  gives 

Ay  =  Au  +  Ay, 


whence 


A?/      Aw      Ay 
Ax      Ax       Ax 


Ay    Au    Av  ,   cfa/    dw    dy 

As  Ax  approaches  zero,  s>  E-  S  approach  ^,  £,  ^ 

respectively.     Therefore 

%        dlfc        dy 

dx      dx      dx 

and  so 

dy  =  du  -\-  dv. 

By  the  same  method  we  can  prove 

d(wdby±iy±  •  •  •  )=<iw±dy=fccfay=b  •  •  •. 

such  that 

Au 

As 

does  not  approach  a  limit  as  Ax  approaches  zero.    Such  a  function  has 
no  derivative  Dxu  and  therefore  no  differential 

du  =  Dxu  dx. 


Chap.  III.  ALGEBRAIC    FUNCTIONS  21 

20.  Proof  of  III.  —  The  differential  of  a  constant  times  a 
function  is  equal  to  the  constant  times  the  differential  of  the 
function. 

Let  y  =  cu. 

Then  y  +  Ay  =  c  (u  +  Au) 

and  so  Ay  =  c  Au, 

Ay        Au 

— -  =  c 

Ax        Ax 

As  Ax  approaches  zero,  -~  and  c-r~m  approach  -j-  and  c-r-- 

—j  x  —j  x  ' ' . c  ax 

Therefore 

dx        dx ' 
whence 

dy  =  c  du. 

Fractions  with  a  constant  denominator  should  be  differen- 
tiated by  this  formula.     Thus 

dl  —  )=  dl—u)—  -du. 

21.  Proof  of  IV.  —  The  differential  of  the  product  of  two 
functions  is  equal  to  the  first  times  the  differential  of  the  second 
plus  the  second  times  the  differential  of  the  first. 

Let  y  =  uv. 

Then  y  +  Ay  =  \u  +  Au)  (p  +  Av) 

=  uv  +  v  Au  +  (u  +  Au)  Av. 

Subtraction  gives 

Ay  =  v  Au  +  (u  +  Au)  Ac, 


whence 


A?/        Au  .    .     .    .    .  Av 
-A—  =  v -.— +  (m .  +  Aw)  x— 
Ax        Ax  y  Ax 


Since  w  is  a  continuous  function,  A?/  approaches  zero  as  Ax 
approaches  zero.     Therefore,  in  the  limit, 

dy        du         dv 
dx        dx         dx' 


22  DIFFERENTIAL   CALCULUS  Chap.  III. 

and  so 

dy  =  v  du  +  u  dv. 

In  the  same  way  we  can  show  that 

d  (uvw)  =  uv  dw  +  uw  dv  +  vw  du. 

22.   Proof  of  V.  —  The  differential  of  a  fraction  is  equal  to 

the  denominator  times  the  differential  of  the  numerator  minus 

the  numerator  times  the  differential  of  the  denominator,   all 

divided  by  the  square  of  the  denominator. 

Let 

u 


u       V 

Then 

and 

A        u  +  Au 

y  +  Ay  =  ■ — ; — t— 
u         J       v  +  Av 

Ay  = 

u  +  Ait      u      v  Au  —  u  Av 

v  +  Av       v         v  (v  +  Av) 

Dividing  by  Ax, 

.           Au         Av 

Av      w  a u-t- 

t-=     Aa;         Ax 
Ax       —, — ; — r-r-« 

v  (v  +  Aw) 

Since  v  is  a  continuous  function  of  x,  Av  approaches  zero  as  Ax 
approaches  zero.     Therefore 

du         dv 


whence 


dy 

dx 

U-rr- 

dx 

dx 

V2 

dy  = 

v  du 

— 

u  dv 

V2 

23.  Proof  of  VI.  —  The  differential  of  a  variable  raised  to  a 
constant  power  is  equal  to  the  product  of  the  exponent,  the  variable 
raised  to  a  power  one  less,  and  the  differential  of  the  variable. 

We  consider  three  cases  depending  on  whether  the  exponent 
is  a  positive  whole  number,  a  positive  fraction,  or  a  negative 
number.  For  the  case  of  irrational  exponent,  see  Ex.  25, 
page  61 . 


Chap.  III.  ALGEBRAIC   FUNCTIONS  23 

(1)  Let  n  be  a  positive  integer  and//  =  un.     Then 
y+i\y  =  (u+Au)n  =  un+nun-1Au+-^-^ — -ft«-2(Aft)2  +    •    •  • 


and 


\y  =  nvr~l  \u  +  n(nQ    ^  un~2  (Aw)2  + 


=  nun~~l. 


Dividing  by  Aft, 

Aft  i      ft  (ft  —  1)       « /  a    \    i 

_ »  =  rmn-1  H >-= — Lun~2  (Aft)  + 

Aft  2 

As  A?*  approaches  zero,  this  approaches 

dy 

du 

Consequently, 

dy  =  nun~l  du. 

2 
ft 

(2)  Let  n  be  a  positive  fraction  -  and  y  =  un  =  uq .     Then 

yq  =  up. 

Since  p  and  q  are  both  positive  integers,  we  can  differentiate 

both  sides  of  this  equation  by  the  formula  just  proved. 

Therefore 

qyq~l  dy  —  rpup~l  du. 
v 

Solving  for  dy  and  substituting  uq  for  y,  we  get 

i  I"1 

du  =  — du  =  -  ft       du  =  ftftn_1  dft. 

p--  y 

§ft      q 

(3)  Let  ft  be  a  negative  number  —m.     Then 

m            ! 
ft  =  ftn   =  ft_m   = 

*  Um 

Since  ra  is  positive,  we  can  find  d  (um)  by  the  formulas  proved 
above.    Therefore,  by  V, 

umd{\)-\d{um)      -mum~xdu  .,  .  , 

du  = M — ^r- - — -  = * =  —mu-m-ldu  =  ftftn_1  du. 

u  (ftm)2  ft2m 


24  DIFFERENTIAL  CALCULUS  Chap.  III. 

Therefore,  whether  n  is  an  integer  or  fraction,  positive  or 
negative, 

d  (un)  =  nun~l  du. 

If  the  numerator  of  a  fraction  is  constant,  this  formula  can  be 
used  instead  of  V.     Thus 


d  [  -  J  =  d  (cur1)  =  —  cu~2  du. 


Exam-pie  1.   y  =  4  x3. 

Using  formulas  III  and  VI, 

dy  =  4  d  (x3)  =  4  •  3  x2  dx  =  12  x2  dx. 

Ex.2,   y  =  Vx  +  -4=  +  3. 

This  can  be  written 

y  =  x*  +  of 5  +  3. 

Consequently,  by  II  and  VI, 

dy      d  (x*)      d  (x~^)      d  (3) 
dx         dx  dx  dx 

1  -idx      1    sdx  .   ~ 

2  dx      2        ax 


2  Vx      2  Vx3 
Ex.3.   i/=  (x  +  a)  (x2-62). 
Using  IV,  with  u  =  x  -{-  a,  v  =  x2  —  b2, 

=  (x  +  a)  (2  x  -  0)  +  (x2  -  b2)  (1  +  0) 
=  3  x2  +  2  ax  -  62. 

Ex.4.  2/  =  -rzr 


Chap.  III.  ALGEBRAIC   FUNCTIONS  25 

Using  V,  with  u =  x2  +  1,  v  =  x2  —  1, 

(x2  -l)d  (x2  +  1)  -  (x2  +  l)d  (x2  -  1) 

dy  = ^zrry 

(x2  -l)2xdx-  (x2+l)2xdx 
(x2  -  l)2 
4xc?x 
(x2  - l)2' 

Ex.  5.   ?/  =  Vx2  -  1. 
Using  VI,  with  u  =  x2  —  1, 

=  1(^-1)^(2,)  =  ^==. 

Ex.  6.   x2  -\-  xy  —  y2  =  1. 

We  can  consider  y  a  function  of  x  determined  by  the  equa- 
tion.    Then 

d  (x2)  +  d  (xy)  -  d  (if)  =  d  (1)  =  0, 
that  is, 

2  x  dx  -\-  x  dy  -\-  y  dx  —  2  y  dy  =  0t 

(2  x  +  y)  dx  +  (x  -  2  y)  dy  =  0. 
Consequently, 

dy      2x  +  y  m 
dx      2  y  —  x 

Ex.7.    x  =  t  +  -,    y  =  t  —  -. 

t  t 

In  this  case 

dx  =  dt  —  jtj  ,     dy  =  dt  +  -g* 
Consequently, 

dy         ^  t2      t2  +  l 
dx  1      t2  -  1 ' 

*2 

Ex.  8.   Find  an  approximate  value  of  y  =  (  1    when 

x  =  0.2.  X  W 


26 


DIFFERENTIAL  CALCULUS 


Chap.  III. 


When  x  =  0,  y  =  1.    Also 


dy  =  - 


2dx 


Z(l-x)Hl+x)$ 


When  #  =  0  this  becomes 

dy  =  —  f  dz. 

If  we  assume  that  d?/  is  approximately  equal  to  A?/,  the  change 
in  y  when  x  changes  from  0  to  0.2  is  approximately 

dy  =  -f  (0.2)  =  -0.13. 

The  required  value  is  then 

y  =  1  -  0.13  =  .87. 


EXERCISES 

In  the  following  exercises  show  that  the  differentials  and  derivatives 
have  the  values  given: 

1.   y  =  3  x4  +  4  x3  -  6  x2  +  5,     dz,   =  12  (x3  +  re2  -  x)  dx. 

d?/      3  a;1-  2 


2.  i/  =  2x5  -  3x3  +  1, 

S3   _  X2  +   1 

3.  y  = ^ , 

4.  ?/  =  (x  +  2  a)  (x  -  a)2, 

5.  y  =  x  (2  x  -  1)  (3  x  +  2), 
1 


dx 


as 


£ 


dy      3x2  -  2a?t-  . 

dx  5 

dz/  =  3  (x2  -  a2)  dx'. 

4-  =  18  x2  +  2  x  -  2. 
ax 


dy  = 


■2xdx 


^   y  ~x2--l'  "*       (x2  +  l)2 

</_  2x  +  3     ,  -22dx        '     d  1 

7'   y=  4x^5'  d*  =  (i^^  10'  dS  e  +  VF=i 


Ve2-  \-d 

V02-  i 


0     d  1    -  2x 

8.   t- 


2x 


dx  (x  -  l)2      (x  -  l)3 

(1  -  t)  dt 


- .      d        . a2  -  2  s2 


12.   d 


2/ 


a2  dy 


9.   d  Vl  +  2*- J2  =    7=      „       - 

T  Vl  +  2*  -  i2  Va2  -  y*      (a2  -  z/2)* 


13.  A(«V«1T^A        _— *^= 
dx  \x  /  £*  Va2  - 

14     ±\f?=l-- 

dxVx*  +  l      (x 


x- 


2X, •  16.  ~xY  =  2xy*  +  2x2*/^. 

(x2  +  l)  Vx4-1  dx  T       *dx 

(2+3X6)"         20  «  / xdx-\-ydy 

16.  d^F-  =  -^(2+3z«M.  17.  d^+!C  =  -^=!' 


■ 

Chap.  III.  ALGEBRAIC  FUNCTIONS  27 

^18.   y  =  (x  +  1)  (2  -  3  x)2  (2  x  -  3)3, 

^  =  (24  +  13x-36x2)(2-3x)(2x  -  3)2. 


19.   y  = 


«'  dx  ™  +  1 


22.    x2  +  i/2  =  a2, 


(a  +  6xn)»  (a  +  6xn)" 

20.  r."bJvy+T,  *-— tL=. 

3  x3  dx      ^4  Vx2  +  1 

(x  +  vT+P)^1  ,   (x  +  vT+T2)"-1 
21-   2/ ^pi + n • 

<fy  =2(x  +  Vl  +  x2)ndx. 
d?/         x 
dx  y 

oo      i   i     i       o  jdy      ay  —  x2 

23.  x3  +  w3  =  3  axw,  ■—  =  -f 

dx      yl  -  ax 

n»     o    o      o         .^9      o  dy      4x  —  3w  —  3 

24.  2  x2  -  3  xy  +  4  w2  =  3  x,  -/  =  — „ ^ • 

'         *  dx  3  x  -  8  2/ 

25.  -  +  -  =  1,  ydx  —  xdy  =  0. 
y      x  *  9 

r>f  1  ,  dx  dl/ 

26'^=^>  '         77r=r=i  +  77r=  =  0. 


x 


Vl  +X4"7"  Vl  +y* 


27.    ?/2n  +  xOT*/n  =  x2m,  mydx  =  nxdz/. 

t  2t  +  S         d*/       _ 

28.  .-j^p   ^  =  T^r^_df  =  5- 

29.   x=<— V<*-1,    y=*+Vp— 1,     xdy  +  ydx  =  0. 

Sat  Sat2  dy      2 1  -  t* 

W'  x  ~  1  +  *3'    y  "  1  +  P'         ete  "  1  -  2*3' 

x 

31.  Given  y  =     /  2      Q>  find  an  approximate  value  for  y  when 

x  =  4.2. 

32.  Find  an  approximate  value  of 


>l. 


X2   —  X  +   1 


X2  +  X  +   1 

when  x  =  .3. 

33.  Given  y  =  x6,  find  d*/  and  Ay  when  x  changes  from  3  to  3.1.  Is 
dy  a  satisfactory  approximation  for  Ay?  Express  the  difference  as  a 
percentage  of  Ay. 

34.  Find  the  slope  of  the  curve 

y  =  x  (x5  +  31)* 
at  the  point  x  =  1. 

35.  Find  the  points  on  the  parabola  y2  =  4  ax  where  the  tangent  is 
inclined  at  an  angle  of  45°  to  the  x-axis. 


28  DIFFERENTIAL  CALCULUS  Chap.  III. 

36.  Given  y  =  (a  +  x)  Va  —  x,  for  what  values  of  x  does  y  increase 
as  x  increases  and  for  what  values  does  y  decrease  as  x  increases? 

37.  Find  the  points  P  (x,  y)  on  the  curve 

,1 

y  =  x  +  - 

where  the  tangent  is  perpendicular  to  the  line  joining  P  to  the  origin. 
,  38.   Find  the  angle  at  which  the  circle 

x2  -\-  y2  =  2x  —  3  y 

intersects  the  £-axis  at  the  origin. 

39.  A  line  through  the  point  (1,  2)  cuts  the  x-axis  at  (x,  0)  and  the 

rlti 

?/-axis  at  (0,  y) .     Find  —  • 

40.  If  x2  —  x  +  2  =  0,  why  is  the  equation 

!c*._,  +  a>-o  • 

not  satisfied? 

41.  The  distances  x,  x'  of  a  point  and  its  image  from  a  lens  are  con- 
nected bv  the  equation 

1  +  1  =  1 

x^x'      /' 

/  being  constant.  If  L  is  the  length  of  a  small  object  extending  along 
the  axis  perpendicular  to  the  lens  and  U  is  the  length  of  its  image,  show 
that 


L       \x) 


approximately,  x  and  x'  being  the  distances  of  the  object  and  its  image 
from  the  lens. 

24.   Higher  Derivatives.  —  The  first  derivative  —  is  a 

function  of  x.     Its  derivative  with  respect  to  x,  written  -7-^ » 

is  called  the  second  derivative  of  y  with  respect  to  x.    That 
is, 

<Py=<3L  (dy\ 

dx2      dx  \dx) 
Similarly, 

d3y       d_  (d2y\ 

dx  \dx2/' 


dx3 
tfy 
dx* 


d  (<Py\ 
=  dx  fei>  etC- 


Chap.  III.  ALGEBRAIC  FUNCTIONS  29 

The  derivatives  of  /  (x)  with  respect  to  x  are  often  written 
/'  (*),  /"  (*),/'"  (*),  etc.     Thus,  if  y  -  /  (*), 

*-r«.  g -/"<*).  3-rw.etc 

Example  1.     y  =  x?. 

Differentiation  with  respect  to  x  gives 

dx4      dx  v  y 

All  higher  derivatives  are  zero. 
Ex.  2.     x2  +  a*/  +  y2  =  1. 
Differentiating  with  respect  to  x, 

whence 

<fy  2x  +  j/| 

dx  x  -\-  2y 

The  second  derivative  is 

o     dy      „ 
d*y  =  _d_  (2x  +  y\  dx~     J 

dx2  ~      dx\x  +  2y)       (x  +  2  ?/)2 

Replacing  -==  by  its  value  in  terms  of  x  and  ?/  and  reducing, 
cix 

d?y  G  (x2  +  xy  +  ?/2)  6 


dx2  (x  +  2  ?/)3  (a  +  2  t/)3 

The  last  expression  is  obtained  by  using  the  equation  of  the 
curve  x2  -\-  xy  +  y2  =  1.  By  differentiating  this  second 
derivative  we  could  find  the  third  derivative,  etc. 


30  DIFFERENTIAL  CALCULUS  Chap.  III. 

25.   Change   of   Variable.  —  We   have   represented   the 

d2v 
second  derivative  by  -~2 .     This  can  be  regarded  as  the  quo- 
tient obtained  by  dividing  a  second  differential 

d2y  =  d  (dy) 

by  (dx)2.    The  value  of  d2y  will  however  depend  on  the  vari- 
able with  respect  to  which  y  is  differentiated. 

d2y 
Thus,  suppose  y  =  x2,  x  =  t3.     Then  v|  =  2  and  so 

d2y  =  2  (dx)2  =  2  (St2  dt)2  =  18  t*  (dt)2. 
If,  however,  we  differentiate  with  respect  to  t,  since  y  =  t&, 

^f  =  30  *4  and 

d2y  =  30  t*  (dt)2, 

which  is  not  equal  to  the  value  obtained  when  we  differen- 
tiated y  with  respect  to  x. 

For  this  reason  we  shall  not  use  differentials  of  the  second 
or  higher  orders  except  in  the  numerators  of  derivatives. 

d2ii  d2ti 

Two  derivatives  like  -^  and  ^  must  not  be  combined  like 

fractions  because  d2y  does  not  have  the  same  value  in  the  two 

cases.  ' 

If  we  have  derivatives  with  respect  to  t  and  wish  to  find 

derivatives  with  respect  to  x,  they  can  be  found  by  using  the 

identical  relation 

du 

d  du  di       dt_  /or\ 

dx  dt  dx      dx 


For  example, 


dt 

d?y 

±(dy\  =  d_(dy\  ctt  =  dt?_ 
dx\dt)      dt\dt)  dx      dx 

dt 


Chap.  III.  ALGEBRAIC  FUNCTIONS  31 

1  1  d2y 

Example.    Given  x  =  t  —  - ,  2/  =  ^  +  7  »  nn^  j~2 ' 

In  this  case 

dy  =  _ £  =  *2-  1 

dz  ::  Jt   .  dt==  t2  +  1* 
dt  +  ¥ 

Consequently, 

dry       d  ft2  -  1\  4  <       d«  __      4 «  1      _        4*3 


1/      V1 


dx2  ' '  dx  \t2  +  1/      (f"  +  l)2  dx     {f  +  I)2  i      (j*  +  1)3 

EXERCISES 
Find  -*  and  -~  in  each  of  the  following  exercises: 

1.  y  =  ?-±4-  v-  6.   x2  +  y2  =  a3. 

a;  —  1 

2.  ?/  =  Vo2  -  x2.  6.   x2  -  2  ?/2  =  1. 

3.  y  =  (x  -  l)3  (x  +  2)4.  7.   xy  =  2  +y. 

4.  ?/2  =  4  x.  8.   xl  +  y*  =  a§. 
9.  If  a  and  6  are  constant  and  y  =  ax2  +  bx,  show  that 

x2  -t^9  -  x  -/  +  bx  =  0. 
dx2  ax 

10.  If  a,  6,  c,  d  are  constant  and  y  =  ax3  +  6x2  +  ex  +  d,  show  that 

Sf-a 

dx^ 

11.  Show  that 

d  I  .dx 


dt  \  dt 
12.    Show  that 


/  .  dx         \      ±d2x 

\ldt-x)  =  t-dr2 


dx 


I   ,d3y       „    A2y  .    _    dy  \        .d*y 

d2u  (Px 

13.  Given     x  =  t2  +  P,     y  =  t2  -  P,  find  ^J  and  — 

14.  By  differentiating  the  equation 

dx      '/./ 
with  respect  to  x,  find  -f\  in  terms  of  derivatives  of  x  with  respect  to  y. 


CHAPTER  IV 
RATES 

26.  Rate  of  Change.  —  If  the  change  in  a  quantity  z  is 
proportional  to  the  time  in  which  it  occurs,  z  is  said  to  change 
at  a  constant  rate.  If  A2  is  the  change  occurring  in  an  in- 
terval of  time  At,  the  rate  of  change  of  z  is 

Az 
At' 

If  the  rate  of  change  of  z  is  not  constant,  it  will  be  nearly 

Az 
constant  if  the  interval  At  is  very  short.     Then  -r-  is  ap- 
proximately the  rate  of  change,  the  approximation  becoming 
greater  as  the  increments  become  less.     The  exact  rate  of 
change  at  the  time  t  is  consequently  defined  as 

Ai=o  At      at 
that  is,  the  rate  of  change  of  any  quantity  is  its  derivative  with 
respect  to  the  time. 

If  the  quantity  is  increasing,  its  rate  of  change  is  positive; 
if  decreasing,  the  rate  is  negative. 

27.  Velocity  Along  a  Straight  Line.  —  Let  a  particle  P 
move  along  a  straight  line  (Fig.  27).     Let  s  =  OP  be  con- 

s  As 

r ..^ 


p 
Fig.  27. 

sidered  positive  on  one  side  of  0,  negative  on  the  other.     If 

the  particle  moves  with  constant  velocity  the  distance  As  in 

the  time  A£,  its  velocity  is 

As 

At* 

If  the  velocity  is  not  constant,  it  will  be  nearly  so  when  A£ 

As 
is  very  short.    Therefore  -r-  is  approximately  the  velocity,  the 

32 


Chap.  IV.  RATES  33 

approximation  becoming  greater  as  A£  becomes  less.     The 
velocity  at  the  time  t  is  therefore  defined  as 

-55B-1-  (27) 

This  equation  shows  that  ds  is  the  distance  the  particle 
would  move  in  a  time  dt  if  the  velocity  remained  constant. 
As  a  rule  the  velocity  will  not  be  constant  and  so  ds  will  be 
different  from  the  distance  the  particle  does  move  in  the 
time  (H. 

When  s  is  increasing,  the  velocity  is  positive;  when  s  is 
decreasing,  the  velocity  is  negative. 

Exam-pie.     A  body  starting  from  rest  falls  approximately 

s  =  16  t2 

feet  in  t  seconds.     Find  its  velocity  at  the  end  of  10  seconds. 
The  velocity  at  any  time  t  is 

v  =  ^  =  32 1  ft./sec.* 

At  the  end  of  10  seconds  it  is 

v  =  320  ft./sec. 

28.  Acceleration  Along  a  Straight  Line.  —  The  accelera- 
tion of  a  particle  moving  along  a  straight  line  is  defined  as 
the  rate  of  change  of  its  velocity.     That  is 

-a-S-  ,   (28) 

This  equation  shows  that  do  is  the  amount  v  would  increase 
in  the  time  dt  if  the  acceleration  remained  constant. 

The  acceleration  is  positive  when  the  velocity  is  increasing, 
negative  when  it  is  decreasing. 

Example.  At  the  end  of  t  seconds  the  vertical  height  of  a 
ball  thrown  upward  with  a  velocity  of  100  ft./sec.  is 

h  =  100  t  -  1G  t\ 

Find  its  velocity  and  acceleration.     Also  find  when  it  is 
rising,  when  falling,  and  when  it  reaches  the  highest  point. 

*  The  notation  ft./sec.  means  feel  per  second.  Similarly,  ft./sec.2, 
used  for  acceleration,  means  feel  per  second  per  second. 


p 


34  DIFFERENTIAL  CALCULUS  Chap.  IV. 

The  velocity  and  acceleration  are 

v  =  ~  =  (100-32  0  ft./sec, 

a  —  -t.  =  —  32  ft./sec.2. 
at 

The  ball  will  be  rising  while  v  is  positive,  that  is,  until  t  — 
^  =  3|.     It  will  be  falling  after  t  =  3|.     It  will  be  at  the 

highest  point  when  t  =  3f . 

29.   Angular  Velocity   and   Acceleration.  —  Consider   a 

body  rotating  about  a  fixed  axis.     Let  0  be  the  angle  turned 

through  at  time  t.  The  angular  veloc- 
ity is  defined  as  the  rate  of  change  of 
d,  that  is, 

angular  velocity  =  co  =  -=r  • 

""  '  The  angular  acceleration  is  the  rate 

of  change  of  angular  velocity,  that  is, 

.  ,  do)      d29 

angular  acceleration  =  a  =  -rr  =  ~r^  • 

Example  1.  A  wheel  is  turning  100  revolutions  per  minute 
about  its  axis.     Find  its  angular  velocity. 

The  angle  turned  through  in  one  minute  will  be 

co  =  100  •  2  7r  =  200  t  radians/min. 

Ex.  2.  A  wheel,  starting  from  rest  under  the  action  of  a 
constant  moment  (or  twist)  about  its  axis,  will  turn  in  t 
seconds  through  an  angle 

6  =  kt\ 

k  being  constant.     Find  its  angular  velocity  and  acceleration 
at  time  t. 
By  definition 

co  =  ,-tt  =  2  kt  rad./sec, 
at 

a  =  -77  =  2  k  rad./sec.2. 
at 


Chap.  IV.  RATES  35 

30.  Related  Rates.  —  In  many  cases  the  rates  of  change 
of  certain  variables  are  known  and  the  rates  of  others  are  to 
be  calculated.  This  is  done  by  expressing  the  quantities 
whose  rates  are  wanted  in  terms  of  those  whose  rates  are 
known  and  taking  the  derivatives  with  respect  to  t. 

Example  1.  The  radius  of  a  cylinder  is  increasing  2  ft. /sec. 
and  its  altitude  decreasing  3  ft./sec.  Find  the  rate  of  change 
of  its  volume. 

Let  r  be  the  radius  and  h  the  altitude.     Then 

V   =  TT2k. 

The  derivative  with  respect  to  t  is 

dv  0dh  ,    _      .  dr 

di  =  vr-it  +  2*rhdt- 

By  hypothesis 

dr  dh  _ 

dt~  Z'     dt~      6' 

Hence 

-r-  =  4  irrh  —  3  7rr2. 
dt 

This  is  the  rate  of  increase  when  the  radius  is  r  and  altitude 
h.     If  r  =  10  ft.  and  h  =  6  ft.,  "~ 

-tt  =  —  60  7r  cu.  ft./sec. 
dt 

Ex.  2.  A  ship  B  sailing  south  at  16  miles  per  hour  is  north- 
west of  a  ship  A  sailing  east  at  10  miles  per  hour.  At  what 
rate  are  the  ships  approaching? 

Let  x  and  y  be  the  distances  of  the  ships  A  and  B  from  the 
point  where  their  paths  cross.  The  distance  between  the 
ships  is  then 

s  =  Vi2  -f-  y2. 

This  distance  is  changing  at  the  rate 

dx         dy 
\  f  ds  ^xTt+y~dt  . 

dt        Vx2  -f- ll2 


t 


r 


36  DIFFERENTIAL  CALCULUS  Chap.  IV. 


By  hypothesis, 

dx       .,_    dy  +„  x  y  At.Q        1 

=  10,  ~  =  —16,      ,  =     ,  =  cos  45    = 


dt  dt  Vx2  +  y2      Vx2  +  y2  V2 

Therefore 

ds      10—16  a;     .   „ 

—  =  — —  =  —  3  V2  mi./nr. 

eft  v2 

The  negative  sign  shows  that  s  is  decreasing,  that  is,  the 
ships  are  approaching. 

EXERCISES 

-  1.  From  the  roof  of  a  house  50  ft.  above  the  street  a  ball  is  thrown 
upward  with  a  speed  of  100  ft.  per  second.  Its  height  above  the  ground 
t  seconds  later  will  be 

h  =  50  +  100  £  -  16  t2. 

Find  its  velocity  and  acceleration  when  t  =  2.     How  long  d^2S  it  con- 
tinue to  rise?     What  is  the  highest  point  reached? 

2.  A  body  moves  in  a  straight  line  according  to  the  law 

s  =  1 1*  -  4  Is  +  16  t\ 

Find  its  velocity  and  acceleration.     During  what  interval  is  the  velocity 
decreasing?     When  is  it  moving  backward? 

3.  If  v  is  the  velocity  and  a  the  acceleration  of  a  particle  moving 
along  the  z-axis,  show  that 

adx  =  v  dv. 

4.  If  a  particle  moves  along  a  line  with  the  velocity 

v2  =  2  gs, 

where  g  is  constant  and  s  the  distance  from  a  fixed  point  in  the  line,  show 
that  the  acceleration  is  constant. 

5.  When  a  particle  moves  with  constant  speed  around  a  circle  with 
center  at  the  origin,  its  shadow  on  the  .r-axis  moves  with  velocity  v 
satisfying  the  equation 

v2  +  n2x2  =  n2r2, 

n  and  r  being  constant.     Show  that  the  acceleration  of  the  shadow  is 
proportional  to  its  distance  from  the  origin. 

6.  A  wheel  is  turning  500  revolutions  per  minute.  What  is  its 
angular  velocity?  If  the  wheel  is  4  ft.  in  diameter,  with  what  speed  does 
it  drive  a  belt? 


Chap.  IV.  RATES  27 

7.  A  rotating  wheel  is  brought  to  rest  by  a  brake.  Assuming  the 
friction  between  brake  and  wheel  to  be  constant,  the  angle  turned 
through  in  a  time  /  will  be 

0  =  a  +  bt  -  cP, 

a,  b,  c  being  constants.     Find  the  angular  velocity  and  acceleration. 
When  will  the  wheel  come  to  rest? 

8.  A  wheel  revolves  according  to  the  law  co  =  30  t  +  /2,  where  co  is 
the  speed  in  radians  per  minute  and  /  the  time  since  the  wheel  started. 
A  second  wheel  turns  according  to  the  law  6  =  \  I2,  where  t  is  the  time 
in  seconds  and  0  the  angle  in  degrees  through  which  it  has  turned.  Which 
wheel  is  turning  faster  at  the  end  of  one  minute  and  how  much? 

9.  A  wheel  of  radius  r  rolls  along  a  line.  If  v  is  the  velocity  and  a 
the  acceleration  of  its  center,  o>  the  angular  velocity  and  a  the  angular 
acceleration  about  its  axis,  show  that 

v  =  rixi,        a  =  rot. 

^0.  The  depth  of  water  in  a  cylindrical  tank,  6  feet  in  diameter,  is 
increasing  1  foot  per  minute.  Find  the  rate  at  which  the  water  is  flow- 
ing in. 

11.  A  stone  dropped  into  a  pond  sends  out  a  series  of  concentric 
ripples.  If  the  radius  of  the  outer  ripple  increases  steadily  at  the  rate 
of  6  ft. /sec,  how  rapidly  is  the  area  of  disturbed  water  increasing  at 
the  end  of  2  seconds? 

12.  At  a  certain  instant  the  altitude  of  a  cone  is  7  ft.  and  the  radius 
of  its  base  3  ft.  If  the  altitude  is  increasing  2  ft. /sec.  and  the  radius  of 
its  base  decreasing  1  ft. /sec,  how  fast  is  the  volume  increasing  or  de- 
creasing? 

13.  The  top  of  a  ladder  20  feet  long  slides  down  a  vertical  wall.  Find 
the  ratio  of  the  speeds  of  the  top  and  bottom  when  the  ladder  makes  an 
angle  of  30  degrees  with  the  ground. 

14.  The  cross  section  of  a  trough  10  ft.  long  is  an  equilateral  triangle. 
If  water  flows  in  at  the  rate  of  10  cu.  ft. /sec,  find  the  rate  at  which  the 
depth  is  increasing  when  the  water  is  18  inches  deep. 

15.  A  man  G  feet  tall  walks  at  the  rate  of  5  feet  per  second  away  from 
a  lamp  10  feet  from  the  ground.  When  he  is  20  feet  from  the  lamp  post, 
find  the  rate  at  which  the  end  of  his  shadow  is  moving  and  the  rate  at 
which  his  shadow  is  growing. 

16.  A  boat  moving  8  miles  per  hour  is  laying  a  cable.  Assuming 
fhat  the  water  is  1000  ft.  deep,  the  cable  is  attached  to  the  bottom  and 
stretches  in  a  straight  line  to  the  stern  of  the  boat,  at  what  rate  is  the 
cable  leaving  the  boat  when  2000  ft.  liave  been  paid  out? 

?J.  Sand  when  poured  from  a  height  on  a  level  surface  forms  a  cone 
wkh  constant  angle  /S  at  the  vertex,  depending  on  the  material.     If  the 


/ 


38  DIFFERENTIAL   CALCULUS  Chap.  [V. 

sand  is  poured  at  the  rate  of  c  cu.  ft. /sec.,  at  what  rate  is  the  radius  in- 
creasing when  it  equals  a? 

18.  Two  straight  railway  tracks  intersect  at  an  angle  of  60  degrees. 
On  one  a  train  is  8  miles  from  the  junction  and  moving  toward  it  at  the 
rate  of  40  miles  per  hour.  On  the  other  a  train  is  12  miles  from  the 
junction  and  moving  from  it  at  the  rate  of  10  miles  per  hour.  Find 
the  rate  at  which  the  trains  are  approaching  or  separating. 

19.  An  elevated  car  running  at  a  constant  elevation  of  50  ft.  above 
the  street  passes  over  a  surface  car,  the  tracks  crossing  at  right  angles. 
If  the  speed  of  the  elevated  car  is  16  miles  per  hour  and  that  of  the  sur- 
face car  8  miles,  at  what  rate  are  the  cars  separating  10  seconds  after 
they  meet? 

20.  The  rays  of  the  sun  make  an  angle  of  30  degrees  with  the  hori- 
zontal. A  ball  drops  from  a  height  of  64  feet.  How  fast  is  its  shadow 
moving  just  before  the  ball  hits  the  ground? 


^ 


CHAPTER  V 
MAXIMA  AND   MINIMA 

31.  A  function  of  x  is  said  to  have  a  maximum  at  x  =  a, 
if  when  x  =  a  the  function  is  greater  than  for  any  other  value 
in  the  immediate  neighborhood  of  a.  It  has  a  minimum  if 
when  x  =  a  the  function  is  less  than  for  any  other  value  of  x 
sufficiently  near  a. 

If  wre  represent  the  function  by  y  and  plot  the  curve 
V  =  f  (x)>  a  maximum  occurs  at  the  top,  a  minimum  at  the 
bottom  of  a  wave. 

If  the  derivative  is  continuous,  as  in  Fig.  31a,  the  tangent 
is  horizontal  at  the  highest  and  lowest  points  of  a  wave  and 
the  slope  is  zero.  Hence  in  determining  maxima  and  minima 
of  a  function  /  (x)  we  first  look  for  values  of  x  such  that 

d 


dx 


f  (*)  =  f  (*)  =  0. 


If  a  is  a  root  of  this  equation,  /  (a)  may  be  a  maximum,  a 
minimum,  or  neither. 


r 

A 

c^y 

^\ 

^s^y 

'    0 

X 

Fig.  31a. 


If  the  slope  is  positive  on  the  left  of  the  point  and  negative 
on  the  right,  as  at  -A,  the  curve  falls  on  both  sides  and  the 
ordinate  is  a  maximum.     That  is,  /  (x)  has  a  maximum  value 

39 


40  DIFFERENTIAL  CALCULUS  Chap.  V. 

at  x  =  a,  iff  (x)  is  positive  for  values  of  x  a  little  less  and  nega- 
tive for  values  a  little  greater  than  a. 

If  the  slope  is  negative  on  the  left  and  positive  on  the  right, 
as  at  B,  the  curve  rises' on  both  sides  and  the  ordinate  is  a 
minimum.  That  is,  /  (x)  has  a  minimum  at  x  =  a,  if  f  (x)  is 
negative  for  values  of  x  a  little  less  and  positive  for  values  a  little 
greater  than  a. 

If  the  slope  has  the  same  sign  on  both  sides,  as  at  C,  the 
curve  rises  on  one  side  and  falls  on  the  other  and  the  ordinate 
is  neither  a  maximum  nor  a  minimum.  That  is,  /  (x)  has 
neither  a  maximum  nor  a  minimum  at  x  =  a  if  f  (x)  has  the 
same  sign  on  both  sides  of  a. 

Example  1.  The  sum  of  two  numbers  is  5.  Find  the  maxi- 
mum value  of  their  product. 

Let  one  of  the  numbers  be  x.     The  other  is  then  5  —  x. 

The  value  of  x  is  to  be  found  such  that  the 

product 

y  =  x  (5  —  x)  =  5  x  —  x2 

is  a  maximum.     The  derivative  is 


(HB. 


^5-2x. 
dx 

This  is  zero  when  x  =  J.  If  x  is  less  than 
Fig.  31b.         t>  the  derivative  is  positive.     If  x  is  greater 

than  f  the  derivative  is  negative.  Near 
x  =  f  the  graph  then  has  the  shape  shown  in  Fig.  31b.  At 
x  =  |  the  function  has  its  greatest  value 

5    fn   5^    _    2  5 

2    W  27    —    "4   • 

Ex.  2.   Find  the  shape  of  the  pint  cup  which  requires  for 
its  construction  the  least  amount  of  tin. 

Let  the  radius  of  base  be  r  and  the  depth  h.     The  area  of 
tin  used  is 

A  =  irr2  +  2  irrh. 

Let  v  be  the  number  of  cubic  inches  in  a  pint.     Then 

v  =  irr2h. 


Chap.  V 


MAXIMA    AND   MINIMA 


41 


(  onsequently, 


and 


i         v 

'l    =   ~2 

irr- 


A  =  7rr2  + 


2v 


Since  -k  and  v  are  constants, 


dr  r2 


/Vr3  —  vy 


This  is  zero  if  irr3  =  v.     If  there  is  a  maximum  or  minimum 
it  must  then  occur  when 


dA 
for,  if  r  has  any  other  value,  -=-  will  have  the  same  sign  on 

both  sides  of  that  value  and  A  will  be  neither  a  maximum  nor 
a  minimum.  Since  the  amount  of  tin  used  cannot  be  zero 
there  must  be  a  least  amount.  This  must  then  be  the  value 
of  A  when  v  =  7T?*3.  Also  v  =  irr2h.  We  therefore  conclude 
that  r  =  h.  The  cup  requiring  the 
least  tin  thus  has  a  depth  equal  to 
the  radius  of  its  base. 

Ex.  3.  The  strength  of  a  rec- 
tangular beam  is  proportional  to 
the  product  of  its  width  by  the 
square  of  its  depth.  Find  the 
strongest  beam  that  can  be  cut 
from  a  circular  log  24  inches  in 
diameter. 

In  Fig.  31c  is  shown  a  section 
of  the  log  and  beam.     Let  x  be  the  breadth  and  y  the  depth 
of  the  beam.     Then 

z2  +  if  =  (24)2. 

The  strength  of  the  beam  is 

S  =  kxif  =  kx  (242  -  x2), 


Fig.  31c. 


42 


DIFFERENTIAL  CALCULUS 


Chap.  V. 


k  being  constant.     The  derivative  of  *S  is 

~  =  k(24:2-3x2). 
dx 

If  this  is  zero,  x  =  ±8  v3.     Since  x  is  the  breadth  of  the 
beam,  it  cannot  be  negative.     Hence 

x  =  sVs 

is  the  only  solution.     Since  the  log  cannot  be  infinitely  strong, 
there  must  be  a  strongest  beam.     Since  no  other  value  can 

give  either  a  maximum  or  a  minimum,  x  =  ^V3  must  be 
the  width  of  the  strongest  beam.     The  corresponding  depth 

is  y  =  8  V6. 

Ex.  4.     Find  the  dimensions  of  the  largest  right  circular 
cylinder  inscribed  in  a  given  right  circular  cone. 

Let  r  be  the  radius  and  h  the  altitude  of  the  cone.     Let 

x  be  the  radius  and  y  the  altitude  of 
an  inscribed  cylinder  (Fig.  31d).  From 
the  similar  triangles  DEC  and  A BC, 

DE      AB 


that  is, 


EC"  BC1 


y 


h 

r 


h  ,  s 

y  =  -  (r  —  x). 


The  volume  of  the  cylinder  is 


Fig.  31d. 


7T 


h 


v  =  irx2y  =  —  (rx2  —  x2) . 


Equating  its  derivative  to  zero,  we  get 

2  rx  -  3  x2  =  0. 

Hence  x  =  0  or  x  =  f  r.  The  value  x  =  0  obviously  does 
not  give  the  maximum.  Since  there  is  a  largest  cylinder,  its 
radius  must  then  be  x  =  f  r.  By  substitution  its  altitude  is 
then  found  to  be  y  =  J  h. 

32.  Method  of  Finding  Maxima  and  Minima.  —  The 
method  used  in  solving  these  problems  involves  the  following 
steps : 


Chap.  V.  MAXIMA  AND   MINIMA  43 


(1)  Decide  what  is  to  be  a  maximum  or  minimum.  Let 
it  be  //. 

(2)  Express  y  in  terms  of  a  single  variable.     Let  it  be  x. 
It  may  be  convenient  to  express  y  temporarily  in  terms  of 

several  variable  quantities.  If  the  problem  can  be  solved  by 
our  present  methods,  there  will  be  relations  enough  to  elimi- 
nate all  but  one  of  these. 

(3)  Calculate  y-  and  find  for  what  values  of  x  it  is  zero. 

(4)  It  is  usually  easy  to  decide  from  the  problem  itself 
whether  the  corresponding  values  of  y  are  maxima  or  minima. 

dv 
If  not,  determine  the  signs  of  -y-  when  #  is  a  little  less  and 

a  little  greater  than  the  values  in  question  and  apply  the 
criteria  given  in  Art.  31. 

EXERCISES 
Find  the  maximum  and  minimum  values  of  the  following  functions: 

1.  2  x2  -  5  x  +  7.  3.   x4  -  2  x2  +  6. 

2.  6  +  12x-x3.  4. 

v  a2  —  x2  ^^> 

Show  that  the  following  functions  have  no  maxima  or  minima: 
.    5.   x3.  7.    G  x5  -  15  x4  +  10  x3./^ 

6.   x"  +  4  x.  8.   x  Va2  +  x1. 

9.    Show  that  x  +  -  has  a  maximum  and  a  minimum  and  that  the 

x 
maximum  is  less  than  the  minimum. 

10.  The  sum  of  the  square  and  the  reciprocal  of  a  number  is  a  mini- 
mum.    Find  the  number. 

11.  Show  that  the  largest  rectangle  with  a  given  perimeter  is  a 
square. 

12.  Show  that  the  largest  rectangle  that  can  be  inscribed  in  a  given 
circle  is  a  square. 

13.  Find  the  altitude  of  the  largest  cylinder  that  can  be  inscribed  in 
a  sphere  of  radius  a. 

14.  A  rectangular  box  with  square  base  and  open  at  the  top  is  to  be 
made  out  of  a  given  amount  of  material.  If  no  allowance  is  made  for 
thickness  of  material  or  waste  in  construction,  what  are  the  dimensions 
of  the  largest  box  that  can  be  made? 


44  DIFFERENTIAL ,  CALCULUS  Chap.  V. 

15.  A  cylindrical  tin  can  closed  at  both  ends  is  to  have  a  given 
capacity.  Show  that  the  amount  of  tin  used  will  be  a  minimum  when 
the  height  equals  the  diameter. 

16.  What  are  the  most  economical  proportions  for  an  open  cylindrical 
water  tank  if  the  cost  of  the  sides  per  square  foot  is  two-thirds  the  cost 
of  the  bottom  per  square  foot? 

17.  The  top,  bottom,  and  lateral  surface  of  a  closed  tin  can  are  to  be 
cut  from  rectangles  of  tin,  the  scraps  being  a  total  loss.  Find  the  most 
economical  proportions  for  a  can  of  given  capacity. 

18.  Find  the  volume  of  the  largest  right  cone  that  can  be  generated 
by  revolving  a  right  triangle  of  hypotenuse  2  ft.  about  one  of  its  sides. 
<-/19.  Four  successive  measurements  of  a  distance  gave  ai}  «2,  «3,  «4  as 
results.  By  the  theory  of  least  squares  the  most  probable  value  of  the 
distance  is  that  which  makes  the  sum  of  the  squares  of  the  four  errors  a 
minimum.     What  is  that  value? 

20.  If  the  sum  of  the  length  and  girth  of  a  parcel  post  package  must 
not  exceed  72  inches,  find  the  dimensions  of  the  largest  cylindrical  jug 
that  can  be  sent  by  parcel  post. 

•  21.  A  circular  filter  paper  of  radius  6  inches  is  to  be  folded  into  a 
conical'  filter.  Find  the  radius  of  the  base  of  the  filter  if  it  has  the 
maximum  capacity. 

22.  Assuming  that  the  intensity  of  light  is  inversely  proportional  to 
the  square  of  the  distance  from  the  source,  find  the  point  on  the  line 
joining  two  sources,  one  of  which  is  twice  as  intense  as  the  other,  at 
which  the  illumination  is  a  minimum. 

23.  The  sides  of  a  trough  of  triangular  section  are  planks  12  inches 
wide.  Find  the  width  at  the  top  if  the  trough  has  the  maximum 
capacity. 

24.  A  fence  6  feet  high  runs  parallel  to  and  5  feet  from  a  wall.  Find 
the  shortest  ladder  that  will  reach  from  the  ground  over  the  fence  to 
the  wall. 

25.  A  log  has  the  form  of  a  frustum  of  a  cone  29  ft.  long,  the  diameters 
of  its  ends  being  2  ft.  and  1  ft.  A  beam  of  square  section  is  to  be  cut 
from  the  log.     Find  its  length  if  the  volume  of  the  beam  is  a  maximum. 

•  26.  A  window  has  the  form  of  a  rectangle  surmounted  by  a  semi- 
circle. If  the  perimeter  is  30  ft.,  find  the  dimensions  so  that  the  greatest 
amount  of  light  may  be  admitted. 

27.  A  piece  of  wire  6  ft.  long  is  to  be  cut  into  6  pieces,  two  of  one  length 
and  four  of  another.  The  two  former  are  bent  into  circles  which  are 
held  in  parallel  planes  and  fastened  together  by  the  four  remaining 
pieces.  The  whole  forms  p,  model  of  a  right  cylinder.  Calculate  the 
lengths  into  which  the  wire  must  be  divided  to  produce  the  cylinder  of 
greatest  volume. 


Chap.  V.  MAXIMA  AND   MINIMA  45 

28.  Among  all  circular  sectors  with  a  given  perimeter,  find  the  one 
which  has  the  greatest  area. 

29.  A  ship  B  is  7">  miles  due  east  of  a  ship  A.  If  B  sails  west  at  12 
miles  per  hour  and  .1  south  at  9  miles,  find  when  the  ships  will  be  closest 
together. 

30.  A  man  on  one  side  of  a  river  h  mile  wide  wishes  to  reach  a  point 
on  the  opposite  side  5  miles  further  along  the  hank.  If  he  can  walk  4 
miles  an  hour  and  swim  2  miles  an  hour,  find  the  route  he  should  take 
to  make  the  trip  in  the  least  time. 

31.  Find  the  length  of  the  shortest  line  which  will  divide  an  equi- 
lateral triangle  into  parts  of  equal  area. 

32.  A  triangle  is  inscribed  in  an  oval  curve.  If  the  area  of  the  tri- 
angle is  a  maximum,  show  graphically  that  the  tangents  at  the  vertices 
of  the  triangle  are  parallel  to  the  opposite  sides. 

33.  A  and  C  are  points  on  the  same  side  of  a  plane  mirror.  A  ray  of 
light  passes  from  A  to  C  by  way  of  a  point  B  on  the  mirror.  Show  that 
the  length  of  the  path  ABC  will  be  a  minimum  when  the  lines  AB, 
CB  make  equal  angles  with  the  perpendicular  to  the  mirror. 

34.  Let  the  velocity  of  light  in  air  be  V\  and  in  water  r>.     The  path  of 

a  ray  of  light  from  a  point  A  in  the  air  to  a  point  C  belowT  the  surface  of 

the  water  is  bent  at  B  where  it  enters  the  water.     If  0i  and  02  are  the 

angles  made  by  AB  and  BC  with  the  perpendicular  to  the  surface,  show 

that  the  time  required  for  light  to  pass  from  A  to  C  will  be  least  if  B  is  so 

placed  that 

sin  0i    _  vi 

sin  02      V2 

35.  The  cost  per  hour  of  propelling  a  steamer  is  proportional  to  the 
cube  of  her  speed  through  the  water.  Find  the  speed  at  which  a  boat 
should  be  run  against  a  current  of  5  miles  per  hour  in  order  to  make  a 
given  trip  at  least  cost. 

36.  If  the  cost  per  hour  for  fuel  required  to  run  a  steamer  is  propor- 
tional to  the  cube  of  her  speed  and  is  S20  per  hour  for  a  speed  of  10  knots, 
and  if  the  other  expenses  amount  to  SI 00  per  hour,  find  the  most  econom- 
ical speed  in  still  water. 

33.  Other  Types  of  Maxima  and  Minima.  —  The  method 
given  in  Art.  31  is  sufficient  to  determine  maxima  and  minima 
if  the  function  and  its  derivative  are  one-valued  and  continu- 
ous. In  Figs.  33a  and  33b  are  shown  some  types  of  maxima 
and  minima  that  do  not  satisfy  these  conditions. 

At  B  and  C,  Fig.  33a,  the  tangent  is  vertical  and  the  de- 
rivative infinite.     At  D  the  slope  on  the  left  is  different  from 


46 


DIFFERENTIAL  CALCULUS 


Chap.  V. 


that  on  the  right.  The  derivative  is  discontinuous.  At  A 
and  E  the  curve  ends.  This  happens  in  problems  where 
values  beyond  a  certain  range  are  impossible.    According  to 


Fig.  33a. 

our  definition,  y  has  maxima  at  A,  B,  D  and  minima  at  C 
and  E. 

If  more  than  one  value  of  the  function  corresponds  to  a 

single  value  of  the  vari- 
able, points  like  A  and 
B,  Fig.  33b,  may  occur. 
At  such  points  two  values 
of  y  coincide. 

These  figures  show 
that  in  determining  max- 
ima and  minima  special 
attention  must  be  given 
to  places  where  the  de- 
Fig.  33b.  rivative  is  discontinuous, 

the  function  ceases  to  exist,  or  two  values  of  the  function 
coincide. 

Example  1.   Find  the  maximum  and  minimum  ordinates 
on  the  curve  yz  =  x2. 

In  this  case,  y  —  x*  and 

dy      2   _i 

— —    =    —  T      3 

dx      3 


No  finite  value  of  x  makes  the  derivative  zero,  but  x  =  0 


Chap.  V. 


MAXIMA  AND  MINIMA 


47 


makes  it  infinite.     Since  y  is  never  negative,  the  value  0  is  a 
minimum  (Fig.  33c). 

Y 


Ex.  2.  A  man  on  one  side  of  a  river  %  mile  wide  wishes  to 
reach  a  point  on  the  opposite  side  2  miles  down  the  river.  If 
he  can  row  6  miles  an  hour  and  walk  4,  find  the  route  he 
should  take  to  make  the  trip  in  the  least  time. 


r--g— i 


V 


Fig.  33d. 


Fig.  33e. 


Let  A  (Fig.  33d)  be  the  starting  point  and  B  the  destina- 
tion.    Suppose  he  rows  to  C,  x  miles  down  the  river.     The 

time  of  rowing  will  be  ^  Vx2  +  J  and  the  time  of  walking 
\  (2  —  x).    The  total  time  is  then 

t  =  i  Vz2  +  i  +  J  (2  -  x). 
Equating  the  derivative  to  zero,  we  get 

1 l  =  n 

6  Vx2  +  \      4 

which  reduces  to  5  x2  +  |  =  0.     This  has  no  real  solution. 


48  DIFFERENTIAL  CALCULUS  Chap.  V. 

The  trouble  is  that  J  (2  —  x)  is  the  time  of  walking  only 
if  C  is  above  B.  If  C  is  below  B,  the  time  is  \  (x  —  2). 
The  complete  value  for  t  is  then 

t  =  ±Vx*  +  l±l(2-x), 

the  sign  being  +  if  x  <  2  and  —  if  x  >  2.  The  graph  of 
the  equation  connecting  x  and  t  is  shown  in  Fig.  33e.  At 
x  =  2  the  derivative  is  discontinuous.  Since  he  rows  faster 
than  he  walks,  the  minimum  obviously  occurs  when  he  rows 
all  the  way,  that  is,  x  =  2. 

EXERCISES 

Find  the  maximum  and  minimum  values  of  y  on  the  following  curves : 

1.  x*  +  y*  =  cfi.  3.    yz  =  x*  —  1. 

2.  y5  =  x2  (x  -  1).  4.    x  =  f  +  t3,  y  =  t2  -  ts. 

5.  Find  the  rectangle  of  least  area  having  a  given  perimeter. 

6.  Find  the  point  on  the  parabola  y2  =  4  x  nearest  the  point 
(-1,0). 

7.  A  wire  of  length  I  is  cut  into  two  pieces,  one  of  which  is  bent  to 
form  a  circle,  the  other  a  square.  Find  the  lengths  of  the  pieces  when 
the  sum  of  the  areas  of  the  square  and  circle  is  greatest. 

8.  Find  a  point  P  on  the  line  segment  AB  such  that  PA2  +  PB2  is 
a  maximum. 

9.  If  the  work  per  hour  of  moving  a  car  along  a  horizontal  track  is 
proportional  to  the  square  of  the  velocity,  what  is  the  least  work  re- 
quired to  move  the  car  one  mile? 

10.    If  120  cells  of  electromotive  force  E  volts  and  internal  resistance 

2  ohms  are  arranged  in  parallel  rows  with  x  cells  in  series  in  each  row, 

the  current  which  the  resulting  battery  will  send  through  an  external 

resistance  of  \  ohm  is 

mxE 


C  = 


x2  +  20 


How  many  cells  should  be  placed  in  each  row  to  give  the  maximum 
current? 


CHAPTER  VI 


DIFFERENTIATION   OF   TRANSCENDENTAL 

FUNCTIONS 

34.   Formulas  for  Differentiating  Trigonometric  Func- 
tions. —  Let  u  be  the  circular  measure  of  an  angle. 


VII. 

VIII. 

IX. 

X. 

XL 
XII. 


d  sin  u    =      cos  u  du. 
d  cos  ti   =  —sin  u  du. 
d  tan  u   =      sec2  u  du. 
d  cot  u   =  —esc2  u  du. 
d  sec  if  =      sec  u  tan  u  du. 
d  esc  u    =  —esc  u  cot  u  du. 


The  negative  sign  occurs  in  the  differentials  of  all  co- 
functions. 

35.    The  Sine  of  a  Small  Angle.  —  Inspection  of  a  table 
of  natural  sines  will   show  that 
the  sine  of  a  small  angle  is  very 
nearly  equal  to  the  circular  meas- 
ure of  the  angle.     Thus 

sin  1°  =  0.017452, 

7T 


180 


=  0.017453. 


We  should  then  expect  that 

..     sin  d       ., 
lim— r—  =  1. 


(35) 


Fig.  35. 


0=0     0 

To  show  this  graphically,  let  0  =  AOP  (Fig.  35).  Draw 
PM  perpendicular  to  OA.  The  circular  measure  of  the  angle 
is  denned  by  the  equation 

arc       arc  AP 


6  = 


rad. 


OP 


49 


50  DIFFERENTIAL  CALCULUS  Chap.  VI. 

Also  sin  6  =  -yp  •     Hence 

sin  6  =     MP     =  chord  QP 
6         arc  AP  '      arc  QP 

As  6  approaches  zero,  the  ratio  of  the  arc  to  the  chord  ap' 

proaches  1  (Art.  53).     Therefore  the  limit  of  — —  is  1. 

u 

36.  Proof  of  VII,  the  Differential  of  the  Sine.  —  Let 

y  =  sin  w. 

Then 

y  +  Aw  =  sin  (w  +  Aw) 

and  so 

Ay  =  sin  (w  +  Aw)  —  sin  w. 

It  is  shown  in  trigonometry-  that 

;     sin  A  -  sin  B  =  2  cos  \  (A  +  B)  sin  i  (A  -  E). 

If  then  A  =  w  +  Aw,  B  =  u, 

Ay  =  2  cos  (w  +  J  Aw)  sin  J  Aw, 
whence 

Aw      _        ,     .   ,  .    x  sin  |  Aw  ,     ,   ,  A    N  sin  \  Aw 

-r^  =  2  cos  (w  +  \  Aw)  -  — r =  cos  (w  +  ^  Aw)  — =-^ — . 

Aw  v        2      y     Aw  _j  v        2       y     |  Aw 

As  Aw  approaches  zero 

sin  J  Aw      sin  0 
JAw     ~  ~0~~ 

approaches  1  and  cos  (w  +  \  Aw)  approaches  cos  w.     There- 
fore 

dw 

-^  =  cos  w. 

aw 

Consequently, 

dy  =  cos  w  aw. 

37.  Proof  of  VIII,  the  Differential  of  the  Cosine. —By 

trigonometry 

cos  u  —  sinf-  —  wj- 


Chap.  VI.  TRANSCENDENTAL    FUNCTIONS  51 

Using  the  formula  just  proved, 

d  cos  u  =  d  sin  (  -  —  u)  =  cosf  -  —  u)  d  f-  —  uj  =  —sin  u  du. 

38.   Proof  of  IX,  X,  XI,  and  XII.    -  Differentiating  both 

sides  of  the  equation 

sin  u 

tan  u  = 

cos  u 

and  using  the  formulas  just  proved  for  the  differentials  of 
sin  u  and  cos  u, 

,  cos  u  d  sin  u  —  sin  u  d  cos  u      cos2  u  du  +  sin2  u  du 

d  tan  U  =  ; = _2  „. 

cos2  u  cos^  u 

=  sec2  u  du. 

By  differentiating  both  sides  of  the  equations 

cos  u  1  1 

cot  u  =  — ,       sec  ^  = ,      esc  u  = , 

sin  u  cos  u  sin  u 

and  using  the  formulas  for  the  differentials  of  sin  u  and  cos  ut 
we  obtain  the  differentials  of  cot  u,  sec  u  and  esc  u. 
Example  1.     y  =  sin2  (x2  +  3). 

Since 

sin2(z2  +  3)  =  [sin  Or2  +  3)  ]2, 

we  use  the  formula  for  u2  and  so  get 

dy  =  2  sin  (x2  +  3)  d  sin  (x2  +  3) 

=  2  sin  02  +  3)  cos  (x2  +  3)  d  02  +  3) 
=  4  x  sin  (z2  +  3)  cos  {x2  +  3)  dx. 

Ex.  2.     ?/  =  sec  2  x  tan  2  x. 

-r  =  sec  2  x  -r-  tan  2  x  +  tan  2  x  -7-  sec  2  x 
dx  dx  dx 

=  sec  2  x  sec2  2  x  (2)  +  tan-2  x  sec  2  x  tan  2  x  (2) 

=  2  sec  2  x  (sec2  2  x  +  tan2  2  x). 

EXERCISES 

In  the  following  exorcises  show  that  the  derivatives  and  differentials 
have  the  values  given: 

.1.   y  =  2  sin  3  x  +  3  cos  2  x,  ^   =  G  (cos  3  x  -  sin  2  x). 

M 


52  DIFFERENTIAL   CALCULUS  Chap.  VI. 

2.  y  =  sin2  - ,  dy  =  sin  -  cos  -  dx. 

3.  y  =  2  cosrcsin2rc  —  sin-re  cos2rc,  dy  =  3  cos  a;  cos  2  re  dx. 
1  —  cos  |  re  dy       1  —  cos  £  re 


4.   y  = 


sin  ^  re  drc         3  sin2  §  x 

dy 


6.    ?/  =  tan  2  re  +  sec  2  re,  —■  =  2  sec  2  re  (sec  2  re  +  tan  2  re). 


drc 

e  4.9X       9X  dy  ,x       -x(      -x  .  ■ 

6.    y  =  cot2 = esc2-,  -p  = —cot- esc2-    esc2- +  cot2; 


d// 


7.   re  =  a  cos  t,  y  =  a  sin3  £,  -p   =  —  3  sin  t  cos  i. 


8.  re  =  a  (<f>  —  sin  </>),?/=  a  (1  —  cos  <£),     y-  =  cot  ^ 

dhi  1 

9.  re  =  cos  £  +  t  sin  t,    y  —  sin  £  —  £  cos  £, 


dx2      t  cos3  £■- 

10.  y—  \  cot5  re  —  ^  cot3  re + cot  re + re,  dy   =  —  cot6xdx. 

11.  y  —  \  tan5  re  +  |  tan3  re  +  tan  re,  dy  =  sec6  re  dx.  , 

12.  u  =  \  sec7  6  —  |  sec5  0  +  i  sec3  0,  c?u  =  tan5  0  sec3  0  d0. 

10  /cos3  re  \   .    1   .   .       .2.  . 

13.  y  =  x  I  — ~ cos  re  I  +  -  sin3  re  -f  5  Sln  ^j     dy  —  x  sin3  re  arc. 

COS  32  /  1  ^  ^)  \  ^) 

14.  y  = ^—  (  -  sin5  x  +  -  sin3  re  +  -  sin  re  j  +  —  re,  dy  =  sin6  re  dx. 

H  _  1  +  sin  re  dy  2  cos  re 

15.  y  =  —    -r-  -  j 


1  —  sin  re  drc        (1  —  sin  re)2 

H «  sec  re  —  tan  re  dy       2  sec  re  (tan  re  —  sec  re) 

16.  y  = — — — j  -^   = — 

sec  re  +  tan  re  dx  sec  re  +  tan  re 

17.  y  =  (cot  re  —  3  tan  re)  Vcotrc,     -~    = =—■ 

dx  2  cot*  re 

18.  If  y  =  A  cos  (nrc)  +  B  sin  (nrc),  where  A  and  B   are  constant, 
show  that 

g  +  *  =  0. 

19.  Find  a  constant  A  such  that  y  =  A  sin  2  re  satisfies  the  equation 

y|  +  5?/  =  3sin2rr. 

-/  20.   Find  constants  A  and  B  such  that  2/  =  A  sin  6  re  +  B  cos  6  re 
.satisfies  the  equation 

21.   Find  the  slope  of  the  curve  1/  =  2  sin  re  +  3  cos  re  at  the  point 

7T 
X    =  —  • 

6 


B 


Chap.  VI.  TRANSCENDENTAL   FUNCTION'S  53 

22.  Find  the  points  on  the  curve  y  —  x  +  sin  2  x  where  the  tangent 
is  parallel  to  the  line  y      2  z  +  3. 

23.  A  weighl  supported  by  a  spring  hangs  at  rest  at  the  origin.    If 

the  weighl  is  lifted  B  distance  A  and  let  fall,  its  height  at  any  subsequent 

time  /  will  be 

y  =  A  cos  (2  irnl), 

n  being  constant.  Find  its  velocity  and  acceleration  as  it  passes  the 
origin.  Where  is  the  velocity  greatest?  Where  is  the  acceleration 
greatest? 

24.  A  revolving  light  5  miles  from  a  straight  shore  makes  one  com- 
plete revolution  per  minute.  Find  the  velocity  along  the  shore  of  the 
beam  of  light  when  it  makes  an  angle  of  60  degrees  with  the  shore  line. 

25.  In  Ex.  24  with  what  velocity  would  the  light  be  rotating  if  the 
spot  of  light  is  moving  along  the  shore  15  miles  per  hour  when  the 
beam  makes  with  the  shore  line  an  angle  of  60  degrees? 

26.  Given  that  two  sides  and  the  included  angle  of  a  triangle  have  at 
a  certain  instant  the  values  6  ft.,  10  ft.,  and  30  degrees  respectively,  and 
that  these  quantities  are  changing  at  the  rates  of  3  ft.,  —2  ft.,  and  10 
degrees  per  second,  how  fast  is  the  area  of  the  triangle  changing? 

27.  O.l  is  a  crank  and  AB  a  connecting  rod  attached  to  a  piston  B 
moving  along  a  line  through  0.  If  OA  revolves  about  0  with  angular 
velocity  a>,  prove  that  the  velocity  of  B  is  coOC,  where  C  is  the  point  in 
which  the  line  BA  cuts  the  line  through  O  perpendicular  to  OB. 

/28.  An  alley  8  ft.  wide  runs  perpendicular  to  a  street  27  ft.  wide. 
What  is  the  longest  beam  that  can  be  moved  horizontally  along  the 
street  into  the  alley? 

29.  A  needle  rests  with  one  end  in  a  smooth  hemispherical  bowl.  The 
needle  will  sink  to  a  position  in  which  the  center  is  as  low  as  possible. 
If  the  length  of  the  needle  equals  the  diameter  of  the  bowl,  what  will  be 
the  position  of  equilibrium? 

30.  A  rope  with  a  ring  at  one  end  is  looped  over  two  pegs  in  the  same 
horizontal  line  and  held  taut  by  a  weight  fastened  to  the  free  end.  If 
the  rope  slips  freely,  the  weight  will  descend  as  far  as  possible.  Find 
the  angle  formed  at  the  bottom  of  the  loop. 

31.  Find  the  angle  at  the  bottom  of  the  loop  in  Ex.  30  if  the  rope  is 
looped  over  a  circular  pulley  instead  of  the  two  pegs. 

y.  32.  A  gutter  is  to  be  made  by  bending  into  shape  a  strip  of  copper  so 
that  the  cross  sect  ion  shall  be  an  arc  of  a  circle.  If  the  width  of  the  strip 
is  a,  find  the  radius  of  the  cross  section  when  the  carrying  capacity  is  a 
maximum. 

33.  A  spoke  in  the  front  wheel  of  a  bicycle  is  at  a  certain  instant  per- 
pendicular to  one  in  the  rear  wheel.  II'  the  bicycle  rolls  straight  ahead, 
in  what  position  will  the  outer  ends  of  the  two  spokes  be  closest  together? 


54  DIFFERENTIAL  CALCULUS  Chap.  VI. 

39.  Inverse  Trigonometric  Functions.  —  The  symbol 
sin-1  x  is  used  to  represent  the  angle  whose  sine  is  x.     Thus 

y  =  sin-1  x,         x  =  sin  y 

are  equivalent  equations.     Similar  definitions  apply  to  cos-1  x 
tan-1  x,  cot-1  x,  sec-1  x,  and  esc-1  x. 

Since  supplementary  angles  and  those  differing  by  mul- 
tiples of  2  7r  have  the  same  sine,  an  indefinite  number  of 
angles  are  represented  by  the  same  symbol  sin-1  x.  The 
algebraic  sign  of  the  derivative  depends  on  the  angle  dif- 
ferentiated. In  the  formulas  given  below  it  is  assumed  that 
sin-1  u  and  esc-1  u  are  angles  in  the  first  or  fourth  quadrant, 
cos-1  u  and  sec-1  u  angles  in  the  first  or  second  quadrant. 
If  angles  in  other  quadrants  are  differentiated,  the  opposite 
sign  must  be  used.  The  formulas  for  tan-1  u  and  cot-1  u 
are  valid  in  all  quadrants. 

40.  Formulas  for  Differentiating  Inverse  Trigonometric 
Functions.  — 

flu 


XIII 

d  sin-1  u   = 

AlXXi 

Vl  -  u2 

XIV 

d  cos-1w    = 

du 

Al  V  • 

Vl  -  u2 

XV. 

d  tan-1  u  = 

du 

1  +  u2 

XVI. 

d  cot-1  u  = 

du 

1  +  u2 

XVII. 

d  sec-1 u  = 

du 

u  Vu2  —  1 

XVIII. 

d  esc-1 u   = 

du 

u  Vu2  —  1 

41.   Proof  of  the  Formulas.  —  Let 

y  =  sin-1  u. 

Then 

sin  y  =  u. 

Differentiation  gives 

cos  y  dy  =  du, 


Chap.  VI.  TRANSCENDENTAL   FUNCTIONS  55 

7  du 

whence  ay  = 

cos  y 

But  

cos  y  =  ±  v  1  —  sin2  y  =  ±  Vl  —  m2. 

If  ?/  is  an  angle  in  the  first  or  fourth  quadrant,  cos  y  is  positive. 

Hence  

cos  y  =  Vl  —  u2 
and  so 

T  du 

dy  = 


Vl  -  u2 
The  other  formulas  are  proved  in  a  similar  way. 

EXERCISES 

3  dx 

1.  y  =  sin-1  (3  x  —  1),  dy  =      .  

V6  x  -  9  x2 

2.  y  =  cos"1  ( 1  -  ~  ),  dy=  dx 

v         a/  V2ax-x2 


4.    y  =  cot  *  I  75  -  7T- 


\2      2x) 


dy        -2 


dx      x2  +  1 
dx 


5.   y  =  sec"1  V4  x  +  1.  dy  =  — 

(4  x  +  1)  Vx 

1  —i       3  dy  1 


6.  y  =  ~  esc 
y       2 

7.  y  =  tan-1 


2  4  x  -  1  dx       V2  +  2  x  -  4  x2 

x  —  a  dy  a 


x  -\-  a 
8.   x  =  esc-1  (sec  0), 


9.    y  =  sin  ' 
10.    y  =  sec-1 


Va2  -  x2 

1 


dx      x2  +  a2 

dx 
d0  ~ 

dy                     a2 

dx       (a2  _  X2)  Va2 

-2x2 

dy             1 

Vl  —  x2  dx  "  Vl  — 


x- 


11.  y  -  |  V«2  -  x2  +  |  sin"^,  ^  =  Vtf  =  tf . 

12.  y  =  tan-i      4  sin  x      >  £      4 

3  +  5  cos  x  </x      5  +  3  cos  x 

13.  y  ^ec-i^TT «  dy  "  -        2 


dx  Vl  -  x2 


56  DIFFERENTIAL  CALCULUS  Chap.  VL 


14.  y  =  a  ain-« -  +  Va2  -  x\        p-  =  Sl0-^- 

y  a  dx       *  a  +  x 

15.  y  =  2  (3  x  +  1)*  +  4  cot"1  £^±i^, 


dy 


<*»      (3x  +  l)*  +  4(3s  +  l)i* 

1ft  if     -i      3-Y  dy   _  1  —  a;2 

lb.    y  --tan     2  +  2;r2'  dx  ==  4  x4  +  17  x2  +  4* 

,x.+  l        2         _,    2x        dy  x  +  1 

17.   y  =  cos-1 — ~ —cos  * 5 •,    -t-= ,  — 

Vx2  —  a2        1  .x      dy  1 

18-    2/  =      o^2 ^73  csc 


x2 


2  a2x2  2  a3  a       dx      ^  Vx2  _  a2 

1 


_,  x  +  Vx2  +  4  x  -  4    dy  _ 
19.    y  =  tan  i-     - >  rf:c  - 


2  rfx      xv/z2+4x-4 

20.  y  =  x  sin-1  x  +  v  1  -  x2  -j4  =     , 

dx2       Vl  -  x2 

21.  y  =  x2  sec"1 1  -  2  Vtf^i,       ^  =  2  a;  sec"1  f;  • 

22.  Let  s  be  the  arc  from  the  x-axis  to  the  point  (x,  y)  on  the  circle 
x2  +  y2  =  a2.     Show  that 

d£  =  _a       ds  =  a      te^M  +  ty. 

dx  y       dy      x 

23.  Let  A  be  the  area  bounded  by  the  circle  x2  +  y2  =  a2,  the  y-axis 

and  the  vertical  line  through  (x,  y).     Show  that 

x 
A  =  xy  +  a2  tan-1  -  ,     dA  =  2  y  dx. 

24.  The  end  of  a  string  wound  on  a  pulley  moves  with  velocity  v 
along  a  line  perpendicular  to  the  axis  of  the  pulley.  Find  the  angular 
velocity  with  which  the  pulley  turns. 

25.  A  tablet  8  ft.  high  is  placed  on  a  wall  with  its  base  20  ft.  above  the 
level  of  an  observer's  eye.  How  far  from  the  wall  should  the  observer 
stand  that  the  angle  of  vision  subtended  by  the  tablet  be  a  maximum? 

42.   Exponential  and  Logarithmic  Functions.  —  If  a  is  a 

positive  constant  and  u  a  variable,  au  is  called  an  exponential 
function.  If  u  is  a  fraction,  it  is  understood  that  au  is  the 
positive  root. 

If  y  =  au,  then  u  is  called  the  logarithm  of  y  to  base  a. 
That  is, 

y  =  au,        u  =  loga'y  (42a) 


Chap.  VI. 


TRANSCENDENTAL  FUNCTIONS 


57 


are   by  definition  equivalent   equations.     Elimination  of   u 
gives  the  important  identity 

alog*u=y.  (42b) 

This  expresses  symbolically  that  the  logarithm  is  the  power 
to  which  the  base  must  be  raised  to  equal  the  number. 

43.   The    Curves   y  =  ax.  —  Let   a   be   greater    than    1. 
The  graph  of 

y  =  ax 


has  the  general  appearance  of  Fig.  43. 
increment  h,  the  increment  in  y 
is 

\y  =  ax+h  —  ax  =  ax  (ah  —  1). 

This  increases  as  x  increases.  If 
then  x  increases  by  successive 
amounts  h,  the  increments  in  y 
form  steps  of  increasing  height. 
The  curve  is  thus  concave  upward 
and  the  arc  lies  below  its  chord. 
The  slope  of  the  chord  AP  join- 
ing A  (0,  1)  and  P  (x,  y)  is 

ax  -  1 

-  • 

x 


If  x  receives  a  small 


Fig.  43. 


As  Pi  moves  toward  A  the  slope 
of  A  Pi  increases.  As  P2  moves  toward  A  the  slope  of  AP2 
decreases.  Furthermore  the  slopes  of  AP2  and  APX  approach 
equality;  for 


a~k-  1 
-k 


=  a 


-k 


'ak  -  X 

k     k 


and  a~k  approaches  1  when  k  approaches  zero.  Now  if  two 
numbers,  one  always  increasing,  the  other  always  decreasing, 
approach  equality,  they  approach  a  common  limit.  Call 
this  limit  m.     Then 


lim 

x=0 


a1 


1 


x 


=  m. 


(43) 


58 


DIFFERENTIAL  CALCULUS 


Chap.  VI. 


This  is  the  slope  of  the  curve  y  =  ax  at  the  point  where  it 
crosses  the  y-axis. 

*  44.   Definition  of  e.  —  We  shall  now  show  that  there  is  a 
number  e  such  that 

lim^^  =  l.  (44) 

x=0       X 

In  fact,  let 


e  =  am 


where  m  is  the  slope  found  in  Art.  43.     Then 


X 


r     ex  -  1      ..     am  -  1       1   ..     am  -  1       1 

lim =  hm =  —  hm =  —  •  m  =  1, 

x=o     x  x=o      ^  m  x±q      x_  m 

m 

The  curves  y  =  ax  all  pass  through  the  point  A  (0,  1). 
Equation  (44)  expresses  that  when  a  =  e  the  slope  of  the 
curve  at  A  is  1.  If  a  >  e  the  slope  is  greater  than  1.  If 
a  <  e,  the  slope  is  less  than  1. 


Y 

a>£ 

i    /                ' 

\    ]    ~          / 

/a=e                    / 

1      /                          ' 

•     /                         / 

/                        ' 

/                       • 

/                      / 

/                     • 

/                    • 

/                   y  a  <  e 

» 

/                / 

/ 

/ 

i 

/              * 

ii 

'             * 

* 

i  / 

*' 

1/ 

*' 

A 

if-'' 

." 

— """^/ 

- — ^-"-" 

0 

X' 

Fig.  44. 

We  shall  find  later  that 

e  =  2.7183 

approximately.  Logarithms  to  base  e  are  called  natural 
logarithms.  In  this  book  we  shall  represent  natural  log- 
arithms by  the  abbreviation  In.  Thus  In  u  means  the  natural 
logarithm  of  u. 


Chap.  VI.  TRANSCENDENTAL  FUNCTIONS  59 

45.  Differentials  of  Exponential  and  Logarithmic  Func- 
tions. — 

XIX.  de*  =  r"  du. 

XX.  <l<iu  =  a"  In  a  du. 

XXI.       <1  In  a  =  —  • 
XXII.      dlog.fi  =  -22 

46.  Proof  of  XIX,  the  Differential  of  eM.  —  Let 

y  —  e". 
Then 

2/  +  Ay  =  eu+Au, 
whence 

A/y  =  eu+Su  —  eu  =  eu  (eAw  —  1) 
and 

Aw  "  Au 

As  Aw  approaches  zero,  by  (44), 

eAu  —  1 
Au 

approaches  1.     Consequently, 

-^  =  eu,        dy  =  eu  du. 
du 

47.  Proof  of  XX,  the  Differential  of  a".  —  The  identity 

a  =  e]na 
gives  au  =  ell]aa. 

Consequently, 

dau  =  eu  ln  a  d  (u  In  a)  =  eu  ln  a  In  a  du  =  a11  In  a  du. 

48.  Proof  of  XXI  and  XXII,  the  Differential  of  a  Log- 
arithm. —  Let 

y  =  \nu. 

Then  ev  =  u. 

Differentiating, 

ev  dy  =  du. 


60  DIFFERENTIAL  CALCULUS  Chap.  VI. 

Therefore 

7        du      du 

u        ey        u 

The  derivative  of  loga  u  is  found  in  a  similar  way. 

Example  1.   y  =  In  (sec2  x). 

d  sec2  x      2  sec  x  (sec  re  tan  x  dx)       _  .  , 

aw  = s—   = ^ =  2  tan  x  dx. 

sec2  x  sec2  a; 

Ex.  2.  y  =  2tan"lz. 

2tan-i*  in  2  da; 


dy  =  2t&n~lx  In  2  d  (tan"1  a;)  = 


1  +  z2 


EXERCISES 


2.   ?/  =  atan2x,  ^  =  2atan2*lnasec22x. 

y  dx      (x  +  l)2 

ez  _  e-x  Wi/        /         9         \2 


4-  ?/  =  ^qp^> 


<  *•  #  =  7F-T— =i»  x:  -  i  .x  i  „-x  i  ■ 


dy  =  f       2       V 
dx      \ex  +  e-a;/ 


6.    y  =  xn  +  nx,  -£  =  nxn_1  +  nx  In  w. 


6.  y  =  axxa,  -£,  =  aV"1  (a  +  xlna). 

7.  y  =  In  (3  x2  +  5  x  +  1), 


dy  6  x  -  -  5 


dx      3  x2  +  5  x  +  1 


8.  2/  =  In  sec2  x,  -v?  =  2  tan  x. 

dx 

9.  i/  =  In  (x  +  Vz2  -  a2), 


dy 
dx 
dy 


dx      Vx2  -  a2 

10.   y  =  In  (sec  ax  +  tan  ax),  --  =  a  sec  ax. 

11    y  =  ln(a*  +  &*)  qj/  =  a*lnq  +  b*ln& 

li.   y       m[fl    -to  ),  dx  aX  __  ba 

i  .10  d?/       cos3x 

12.    w  =  In  sm  x  +  2  cos2  x,  -p-  =  — 

y  dx       smx 

1 .  x       1  cos  x  dy  1 

^2  2      2  sin2  x  dx      sin3  x 


Y* 


Chap.  VI.  TRANSCENDENTAL  FUNCTIONS  61 

1      ■'      4     x2-4      x2-4'         dx      x(x2-4)2 
15.    j/  =  -  In 


a      a  +  Va2  -  x2  dx       x  Va2  - 


.(•- 


16.    ,/=ln(vg+3  +  Vx  +  2)+V(x  +  3)(x  +  2),   ^  =  \j |±|. 


17.    ?/  =  In  (V.c  +  a  +  Vx), 


dx       V  x  +  2 

ft=-         1 

dx      2  Vx2  +  ax 

18.  y  =x  tan"1  --£  In  (x2+a2),      ft  =  tan"1  -  • 
J  a     2  dx  a 

(111 

19.  y  =  cax  (sin  ax  —  cos  ax).         ~  =  2  aeox  sin  ax. 

r  dx 

a*?/ 

20.  ?/  =  j  tan4  x  —  ^  tan2  x  —  In  cos  x,        ~  =  tan5  x. 

22.  x  =  e<  +  e-*,   7/  =  6*  -  e-«,      ^  =  --,• 

23.  2/^ln,,  g=J.(21nx-3). 

dnv 

24.  y  =  xe1,  -t-|  =  (x  +  n)  e*. 

25.  By  taking  logarithms  of  both  sides  of  the  equation  y  =  xn  and 
differentiating,  show  that  the  formula 

d 
dx 
is  true  even  when  n  is  irrational. 

26.  Find  the  slope  of  the  catenary 

y  =  \w  +  e~~a) 

at  x  =  0. 

.  27.    Find  the  points  on  the  curve  y  =  c2X  sin  x  where  the  tangent 
is  parallel  to  the  x-axis. 

28.  If  y  =  Aenx  +  J3e_nx,  where  A  and  B  are  constant,  show  that 

dx2       n  y       U' 

29.  If  y  —  ze-31,  where  z  is  any  function  of  x,  show  that 

dx2         dx         ■  dx2 

30.  For  what  values  of  x  does 
y  =  5  In  (x  -  2)  +  3  In  (x  +  2)  +  4  x 

increase  as  x  increases? 


-  xn  =  nx"-1 


62  DIFFERENTIAL   CALCULUS  Chap.  VI. 

31.  From  equation  (44)  show  that 

e  =  lim  (1  +  x)*- 

x=0 

32.  If  the  space  described  by  a  point  is  s  =  ael  +  be'*,  show  that  the 
acceleration  is  equal  to  the  space  passed  over. 

33.  Assuming  the  resistance  encountered  by  a  body  sinking  in  water 
to  be  proportional  to  the  velocity,  the  distance  it  descends  in  a  time  t  is 

«-|* +  &(«"*-«. 

g  and  k  being  constants.     Show  that  the  velocity  v  and  acceleration  a 

satisfy  the  equation 

a  =  g  —  kv. 

Also  show  that  for  large  values  of  t  the  velocity  is  approximately  con- 
stant. 

34.  Assuming  the  resistance  of  air  proportional  to  the  square  of  the 
velocity,  a  body  starting  from  rest  will  fall  a  distance 


9  ]n(ck*+e-M\ 
~/c2      V        2         ) 


in  a  time  t.     Show  that  the  velocity  and  acceleration  satisfy  the  equa- 
tion 

k2v2 

a  =  g 

y         9 

Also  show  that  the  velocity  approaches  a  constant  value. 


CHAPTER  VII 


Fig.  49. 


GEOMETRICAL   APPLICATIONS 

49.  Tangent  Line  and  Normal.  —  Let  Wi  be  the  slope 
of  a  given  curve  at  Pi  (xu  2/i).  It  is  shown  in  analytic 
geometry  that  a  line 
through  (xi,  y\)  with  slope 
vii  is  represented  by  the 
equation 

y  —  yi  =  Wi  (x  —  x{). 

This  equation  then  rep- 
resents the  tangent  at 
(#i,  V\)  where  the  slope  of 
the  curve  is  wii. 

The  line  PiN  perpen- 
dicular to  the  tangent  at 
its  point  of  contact  is 
called  the  normal  to  the  curve  at  Pi.     Since  the  slope  of  the 

tangent  is  mi,  the  slope  of  a  perpendicular  line  is and  so 

»-»- -£&-*) 

is  the  equation  of  the  normal  at  fa,  yi). 

Example  1.   Find  the  equations  of  the  tangent  and  normal 
to  the  ellipse  x2  +  2  y2  =  9  at  the  point  (1,2). 
The  slope  at  any  point  of  the  curve  is 

dy  x 

dx  2y 

At  (1,  2)  the  slope  is  then 

mi=  -\. 

The  equation  of  the  tangent  is 

y-2  =  -l(x-l), 

63 


64 


DIFFERENTIAL  CALCULUS 


Chap.  VII. 


and  the  equation  of  the  normal  is 

y-2  =  4(x-l). 

Ex.  2.   Find  the  equation  of  the  tangent  to  x2  —  y2  =  a2 
at  the  point  (xh  2/1). 

Xi 

The  slope  at  (xi,  yi)  is  — .     The  equation  of  the  tangent 


is  then 


which  reduces  to 


2/i 


y  -  2/1  =  -  0 
2/1 


Xi) 


Xix  -  yiy  =  xi2  -  yi2. 


Since  fa,  yi)  is  on  the  curve,  xi1  —  yi2  =  a2.     The  equation 
of  the  tangent  can  therefore  be  reduced  to  the  form 

xix  -  yiy  =  a2. 

50.  Angle  between  Two  Curves.  —  By  the  angle  be- 
tween two  curves  at  a 
point  of  intersection  we 
mean  the  angle  between 
their  tangents  at  that 
point. 

Let  mi  and  m2  be  the 
slopes  of  two  curves  at 
a  point  of  intersection. 
It  is  shown  in  analytic 
geometry  that  the  angle 
0  from  a  line  with  slope 
mi   to   one  with   slope 


0 


Fig.  50a. 
m2  satisfies  the  equation 


tan  /3  = 


ra2  —  mi 
1  +  ?ftim2 


(50) 


This  equation  thus  gives  the  angle  (3  from  a  curve  with  slope 
mi  to  one  with  slope  m2,  the  angle  being  considered  positive 
when  measured  in  the  counter-clockwise  direction. 


Chap.  VII. 


GEOMETRICAL  APPLICATIONS 


65 


Example.     Find  the  angles  determined  by  the  line  y  =  x 
and  the  parabola  y  =  x2. 

Solving  the  equations  simultaneously,  we  find  that  the  line 
and     parabola     intersect    at 
(1,  1)  and  (0,  0).     The  slope 
of  the  line  is  1.     The  slope  at 
any  point  of  the  parabola  is 

-y-  =  2x. 
ax 


At  (1,  1)  the  slope  of  the 
parabola  is  then  2  and  the 
angle  from  the  line  to  the 
parabola  is  then  given  by 


Fig.  50b. 


tan  (3i  = 


1 


1 
3' 


whence 


1  +  2 

0i  =  tan-1  J  =  18°  26'. 
At  (0,  0)  the  slope  of  the  parabola  is  0  and  so  the  angle 


from  the  line  to  the  parabola  is  given  by  the  equation 

0-  1 


tan  j32  = 


=  -1, 


whence 


1+0 

&  =  -45°. 

The  negative  sign  signifies  that  the  angle  is  measured  in  the 
clockwise  direction  from  the  line  to  the  parabola. 

EXERCISES 

Find  the  tangent  and  normal  to  each  of  the  following  curves  at  the 
point  indicated:  v 

1.  The  circle  x2  +  if  =  5  at  (-1,  2). 

2.  The  hyperbola  xy  =  4  at  (1,  4). 

3.  The  parabola  if  -  ax  at  x  =  a. 

4.  The  exponential  curve  y  =  abx  at  x  =  0. 

5.  The  sine  curve  y  =  3  sin  x  at  x 

r-1 


7T 

6" 


6.    The  ellipse  --,  +  ]jj>  =  t,  at  {xu  ffi). 


66  DIFFERENTIAL  CALCULUS  Chap.  VII. 

7.  The  hyperbola  x2  +  xy  —  y2  =  2  x,  at  (2, 0). 

8.  The  semicubical  parabola  yz  =  x2,  at  (  —  8,  4). 

9.  Find  the  equation  of  the  normal  to  the  cycloid 

x  =  a  (0  —  sin  <j>),         y  =  a  (1  —  cos  0) 

at  the  point  <£  =  <j>\.  Show  that  it  passes  through  the  point  where  the 
rolling  circle  touches  the  x-axis. 

Find  the  angles  at  which  the  following  pairs  of  curves  intersect: 

10.  y2  =  4  x,     x2  =  4  y.  13.    y  =  sin  x,     y  =  cos  x. 

11.  x2  +  y2  =  9,  x2  +  y2  —  6  x  =  9.   14.   ?/  =  logio  £,     ?/  =  In  x. 

12.  x2  -}-  y2  +  2  s  =  7,    i/2=4r     15.    */  =  \  (ex  4-  e""*),     y  =  2ex. 

16.  Show  that  for  all  values  of  the  constants  a  and  b  the  curves 

x2  —  y2  =  a2,        xy  =  b2  * 

intersect  at  right  angles. 

17.  Show  that  the  curves 

y  =  e®*,         y  =  eax  sin  (bx  -4-  c) 

are  tangent  at  each  point  of  intersection. 

18.  Show  that  the  part  of  the  tangent  to  the  hyperbola  xy  =  a2  in- 
tercepted between  the  coordinate  axes  is  bisected  at  the  point  of  tan- 
gency. 

19.  Let  the  normal  to  the  parabola  y2  =  ax  at  P  cut  the  x-axis  at  N. 
Show  that  the  projection  of  PN  on  the  a>axis  has  a  constant  length. 

20.  The  focus  F  of  the  parabola  y2  =  ax  is  the  point  (|  a,  0).  Show 
that  the  tangent  at  any  point  P  of  the  parabola  makes  equal  angles 
with  FP  and  the  line  through  P  parallel  to  the  axis. 

21.  The  foci  of  the  ellipse 

^  +  %  =  1,         a  >  b 

a2      b2 

are  the  points  F'  (  -  vV  -  b2,  0)  and  F  (Va2  -  b2,  0) .  Show  that  the 
tangent  at  any  point  P  of  the  ellipse  makes  equal  angles  with  FP  and 
F'P. 

(x  x\ 

ea  +e  a  J ,  M  the 

projection  of  P  on  the  a>axis,  and  N  the  projection  of  M  on  the  tangent 
at  P.     Show  that  MN  is  constant  in  length. 

23.   Show  that  the  portion  of  the  tangent  to  the  tractrix 


a  .    la  -\-  Va 

=  2ln "3=' 

*      \a  —  v  a 


-  x2 


y  =  ;  — -         =l  -  Va2  -  x2 

*      \a  —  v  a2  —  x2/ 

intercepted  between  the  y-axis  and  the  point  of  tangency  is  constant  in 
length. 


(Hap.  VII.  GEOMETRICAL  APPLICATIONS  67 

24.  Show  that  the  angle  between  the  tangent  at  any  point  P  and  the 
line  joining  P  to  the  origin  is  the  same  at  all  points  of  the  curve 

In  Vx2  +  if  =  k  tan"1  ^  • 

J  x 

25.  A  point  at  a  constant  distance  along  the  normal  from  a  given 
curve  generates  a  curve  which  is  called  parallel  to  the  first.  Find  the 
parametric  equations  of  the  parallel  curve  generated  by  the  point  at 
distance  h  along  the  normal  drawn  inside  of  the  ellipse 

x  =  a  cos  <j>,         y  =  b  sin  </>. 

51.  Direction  of  Curvature.  —  A  curve  is  said  to  be  con- 
cave upward  at  a  point  P  if 
the  part  of  the  curve  near  P 
lies  above  the  tangent  at  P. 
It  is  concave  downward  at  Q 
if  the  part  near  Q  lies  below 
the  tangent  at  Q. 

d"y 
At  points  where  ~r\  is  pos-  Fig.  51. 

itive,  the  curve  is  concave  upward;  where  3-5  is  negative,  the 

curve  is  concave  downward. 
For 
I  d?y  =  d  fdy\ 

dx2      dx  \dx) 

d?y  .  dy 

If  then  -~  is  positive,  by  Art.  13,  ~ ,  the  slope,  increases  as  x 

increases  and  decreases  as  x  decreases.     The  curve  therefore 

rises  above  the  tangent   on   both  sides  of   the  point.     If, 

d"if 
however,  -t4  is  negative,  the  slope  decreases  as  x  increases 

and  increases  as  x  decreases,  and  so  the  curve  falls  below  the 
tangent. 

52.  Point  of  Inflection.  —  A  point  like  A  (Fig.  52a),  on 
one  side  of  which  the  curve  is  concave  upward,  on  the  other 
concave  downward,  is  called  a  point  of  inflection.  It  is 
assumed  that  there  is  a  definite  tangent  at  the  point  of  in- 
flection.    A  point  lik?  B  is  not  called  a  point  of  inflection. 


68 


DIFFERENTIAL  CALCULUS 


Chap.  VLi 


The  second  derivative  is  positive  on  one  side  of  a  point  o  *■■ 

inflection  and  negative  on  the  other.     Ordinary  function:  (&' 

change  sign  only  by  passing  through  zero  or  infinity.     Henc(i  i 

d2y  .  in 

to  find  points  of  inflection  we  find  where  i-^  is  zero  or  infinite 


rJ- 


Fig.  52a. 


If  the  second  derivative  changes  sign  at  such  a  point,  it  is  a 
point  of  inflection.  If  the  second  derivative  has  the  same 
sign  on  both  sides,  it  is  not  a  point  of  inflection. 


Fig.  52b. 


0 


Fig.  52c. 


Example  1.   Examine  the  curve  3  y  =  x4  —  6  x>  for  direc- 
tion of  curvature  and  points  of  inflection. 
The  second  derivative  is 

dhj 


dx2 


=  4  (x2  -  1). 


'"'II 


Chap.  VII.  GEOMETRICAL  APPLICATIONS  69 


t 


of  This  is  zero  at  x  —  =b  1.  It  is  positive  and  the  curve  con- 
3Ds  cave  upward  on  the  left  of  x  =  —1  and  on  the  right  of 
icbJjc  =  +1.     It  is  negative  and  the  curve  concave  downward 

between    x  =  —  1    and    x  =  +  1.     The    second    derivative 

changes  sign  at  A  (  —  1,  — $)   and  J5  (+1,  —  §),  which  are 

therefore  points  of  inflection  (Fig.  52b). 

Ex.  2.    Examine  the  curve  y  =  x*  for  points  of  inflection. 

In  this  case  the  second  derivative  is 

This  is  zero  when  x  is  zero  but  is  positive  for  all  other  values 
)of  x.     The  second  derivative  does  not  change  sign  and  there 
is  consequently  no  point  of  inflection  (Fig.  52c). 

Ex.  3.   If  x  >  0,  show  that  sin  x  >  x  —  —  •* 

Let 

a? 

y  =  sm  x  —  x  +  —. i  • 
u  3! 

We  are  to  show  that  y  >  0.     Differentiation  gives 

dy  -    ,  #2       d2y 

~Y~  =  COS  X  —  1  +  jr.  ,       ~  =   —  sin  £  +  #. 

ax  2!       ax2 

d?y 
When  x  is  positive,  sin  x  is  less  than  x  and  so  -t4  is  positive. 


d-x1 

Therefore  -~  increases  with  x.     Since  -~  is  zero  when  x  is 
ax  dx 


^2/ 


zero,  ~  is  then  positive  when  x  >  0,  and  so  ?/  increases  with 

x.     Since  2/  =  0  when  x  =  0,  y  is  therefore  positive  when 
x  >  0,  which  was  to  be  proved. 

*  If  n  is  any  positive  integer  n !  represents  the  product  of  the  integers 
from  1  to  n.     Thus 

3!  =  1-2-3  =  6. 


70  DIFFERENTIAL  CALCULUS  Chap.  VII. 


EXERCISES 

Examine  the  following  curves  for  direction  of  curvature  and  points 
of  inflection: 

1.  y  =  xz  —  3  x  +  3.  5.  y  =  xex. 

2.  y  =  2  x*  -  3  x2  -  6  x  +  1.  6.  y  =  e~x\ 

3.  y  =  x*  -  4  x3  +  6  x2  +  12  z.  7.  z2?/  -  4  z  +  3  y  =  0. 

4.  y3  =  x  —  1.  8.  as  =  sin  i,     ?/  =  £  sin  3  t. 

Prove  the  following  inequalities: 

9.   x  In  x  -  x  -  |-  +  |  >  0,      if  0  <  a  <  1. 

10.  (x  -  1)  ex  +  1  >  0,  if  a;  >  0. 

n 

11.  e*  <  1  +  x  +  ^  ea,  if  0  <  x  <  a. 

X2  IT  IT 

12.  In  sec  x  >  —  ,  if  —  -  <  x  <  -  • 

13.  According  to  Van  der  Waal's  equation,  the  pressure  p  and 
volume  v  of  a  gas  at  constant  temperature  T  are  connected  by  the  equa- 
tion 

RT  a 


P  = 


m  (v  —  b)       v 


•>  t 


a,  b,  m,  and  R  being  constants.  If  p  is  taken  as  ordinate  and  v  as  ab- 
scissa, the  curve  represented  by  this  equation  has  a  point  of  inflection. 
The  value  of  T  for  which  the  tangent  at  the  point  of  inflection  is  hori- 
zontal is  called  the  critical  temperature.  Show  that  the  critical  tem- 
perature is 

„       8  am 

53.  Length  of  a  Curve.  —  The  length  of  an  arc  PQ  of  a 
curve  is  denned  as  the  limit  (if  there  is  a  limit)  approached 
by  the  length  of  a  broken  line  with  vertices  on  PQ  as  the 
number  of  its  sides  increases  indefinitely,  their  lengths  ap- 
proaching zero. 

We  shall  now  show  that  if  the  slope  of  a  curve  is  continu- 
ous the  ratio  of  a  chord  to  the  arc  it  subtends  approaches  1  as 
the  chord  approaches  zero. 

In  the  arc  PQ  (Fig.  53)  inscribe  a  broken  line  PABQ. 
Projecting  on  PQ,  we  get 

PQ  =  proj.  PA  +  proj.  AB  +  proj.  BQ. 


Chap.  VII. 


GEOMETRICAL  APPLICATIONS 


71 


The  projection  of  a  chord,  such  as  AB,  is  equal  to  the  prod- 
uct of  its  length  by  the  cosine  of  the  angle  it  makes  with  PQ. 
On  the  arc  AB  is  a  tangent  RS  parallel  to  AB.  Let  a  be  the 
largest  angle  that  any  tangent  on  the  arc  PQ  makes  with  the 

Y 


o  x 

Fig.  53. 

chord  PQ.  The  angle  between  RS  and  PQ  is  not  greater 
than  a.  Consequently,  the  angle  between  AB  and  PQ  is  not 
greater  than  a.     Therefore 

proj.  AB  =  AB  cos  a. 
Similarly, 

proj.  PA  =  PA  cos  a, 

proj.  BQ  =  BQ  cos  a. 

Adding  these  equations,  we  get 

PQ  =  (PA  +  AB  +  BQ)  cos  a. 

It  is  evident  that  this  result  can  be  extended  to  a  broken 
line  with  any  number  of  sides.  As  the  number  of  sides  in- 
creases indefinitely,  the  expression  in  parenthesis  approaches 
the  length  of  the  arc  PQ.     Therefore 

PQ  =  arc  PQ  cos  a, 
that  is, 

chord  PQ 

^r^—  =  cos  a. 

arc  PQ 

If  the  slope  of  the  curve  is  continuous,  the  angle  a  ap- 
proaches zero  as  Q  approaches  P.     Hence  cos  a  approaches 


1  and 


..      chord  PQ 

hm 57^-  =1. 

q=p    arc  PQ 


72 


DIFFERENTIAL   CALCULUS 


Chap.  VII. 


Since  the  chord  is  always  less  than  the  arc,  the  limit  cannot 
be  greater  than  1.     Therefore,  finally, 


,.      chord  PQ      . 
hm ^^  =  1. 


(53) 


q=p    arc  PQ 

54.  Differential  of  Arc.  —  Let  s  be  the  distance  measured 
along  a  curve  from  a  fixed  point  A  to  a  variable  point  P. 
Then  s  is  a  function  of  the  coordinates  of  P.  Let  4>  be  the 
angle  from  the  positive  direction  of  the  a>axis  to  the  tangent 
PT  drawn  in  the  direction  of  increasing  s. 


o 


o 


Fig.  54a. 


Fig.  54b. 


If  P  moves  to  a  neighboring  position  Q,  the  increments  in 
x,  y,  and  s  are 

Ax  =  PR,        Ay  =  RQ,        As  =  arc  PQ. 
From  the  figure  it  is  seen  that 

cos  (RPQ)  = 


sin  (RPQ)  = 


PQ'  As  PQ' 

Ay      Ay  As 


PQ      As  PQ 

As  Q  approaches  P,  RPQ  approaches  cf>  and 

As         arc  PQ 
PQ  ~  chord  PQ 

approaches  1.     The  above  equations  then  give  in  the  limit 


dx 


.     ,       dy 

sin  d>  =  -~ 

ds 


(54a) 


Chap.  VII. 


GEOMETRICAL   APPLICATIONS 


73 


These  equations  express  thai  dx  and  dy  are  the  sides  of  a 
right  triangle  with  hypotenuse  ds  extending  along  the  tangent 
(Fig.  54b).  All  the  equations  connecting  dx,  dy,  ds,  and  <f>  can 
be  read  off  this  triangle.     One  of  particular  importance  is 

ds2  =  dx2  +  dif.  (54b) 

55.  Curvature.  —  If  an  arc  is  everywhere  concave  toward 
its  chord,  the  amount  it  is  bent  can  be  measured  by  the  angle 
/3  between  the  tangents  at  its  ends.     The  ratio 

ft  4>'  -  <f>      A<£ 

arc  PP'  ~      As     ~  As 

is  the  average  bending  per  unit  length  along  PPf.     The 
limit  as  P'  approaches  P, 

&<t>      d<j> 
ds 


lim 

As=0  AS 


is  called  the  curvature  at  P.     It  is  greater  where  the  curve 
bends  more  sharply,  less  where  it  is  more  nearly  straight. 


Fig.  55a. 

Fig.  55b 

In  case  of  a  circle  (Fig.  55b) 

*=*+!> 

s  =  ad. 

Consequently, 

d<f>        dB 

1 

<ls       add 

—     » 

a 

74  DIFFERENTIAL  CALCULUS  Chap.  VII. 

that  is,  the  curvature  of  a  circle  is  constant  and  equal  to  the 

reciprocal  of  its  radius. 

56.   Radius  of  Curvature.  — We  have  just  seen  that  the 

radius  of  a  circle  is  the  reciprocal  of  its  curvature.     The 

radius  of  curvature  of  any  curve  is  defined  as  the  reciprocal  of 

its  curvature,  that  is, 

ds 
radius  of  curvature  =  p  =  -j-  •  (56a) 

aq> 

It  is  the  radius  of  the  circle  which  has  the  same  curvature  as 
the  given  curve  at  the  given  point. 

To  express  p  in  terms  of  x  and  y  we  note  that 

<j>  =  tan-1  -y-  • 
dx 

Consequently, 

d^-dx 

d<f>  = — L—dfdy^-    dx* 


Also  ds  =  Vdx2  +  dy2. 

Substituting  these  values  for  ds  and  d<f>,  we  get 

MIT 

p  =  — 


dx2 


(56b) 


If  the  radical  in  the  numerator  is  taken  positive,  p  will  have 

d2y 
the  same  sign  as  -r\ ,  that  is,  the  radius  will  be  positive  when 

the  curve  is  concave  upward.     If  merely  the  numerical  value 
is  wanted,  the  sign  can  be  omitted. 
By  a  similar  proof  we  could  show  that 


til" 

P"  d^x 

dy2 


-i  3 

2 


(56c) 


Chap.  VII.  GEOMETRICAL  APPLICATIONS  75 

Example  1.   Find  the  radius  of  curvature  of  the  parabola 
y2  =  4  x  at  the  point  (4,  4). 

At  the  point  (4,  4)  the  derivatives  have  the  values 


dij    .21 

dx      y      2J 
Therefore 

*V          4 

dx8          y3 

1 
16 

[> + (D7 

dx2 

(■ + S' 

1 

16 

-  io- 

The  negative  sign  shows  that  the  curve  is  concave  downward. 

The  length  of  the  radius  is  10  V5. 

Ex.  2.   Find  the  radius  of  curvature  of  the  curve  repre- 
sented by  the  polar  equation  r  =  a  cos  0. 

The  expressions  for  x  and  y  in  terms  of  0  are 

x  =  r  cos  0  =  a  cos  0  cos  6  =  a  cos2  0, 
y  =  r  sin  6  =  a  cos  6  sin  d. 

Consequently, 

dy      a  (cos2  0  —  sin2  0)        a  cos  2  0 


eta       —  2  a  cos  0  sin  0       —  a  sin  2  0 


=  -cot  2  0, 


cP|/         \dx)  _      2  esc2  2  0  rffl  =       2      3  2 
dx2         dx  a  sin  2  0  d0  a 

_   [1  +  cot2  2  0]*  _  _     (esc2  2  0)^  =  _a 

p  2       ,n/1  a  2  esc3  2  0  =        2* 

esc3  2  0 

a 

The  radius  is  thus  constant.     The  curve  is  in  fact  a  circle. 

57.  Center  and  Circle  of  Curvature.  —  At  each  point  of 
a  curve  is  a  circle  on  the  concave  side  tangent  at  the  point 
with  radius  equal  to  the  radius  of  curvature.  This  circle  is 
called  the  circle  of  curvature.  Its  center  is  called  the  center 
of  curvature. 

Since  the  circle  and  curve  are  tangent  at  P,  they  have  the 


76 


DIFFERENTIAL  CALCULUS 


Chap.  VII. 


dy 
same  slope  —■  at  P.     Since  they  have  the  same  radius  of 

curvature,  the  second  derivatives  will  also  be  equal  at  P. 


Fig.  57. 


The  circle  of  curvature  is  thus  the  circle  through  P  such  that 

dy         d^y 

-j-  and  7-5  have  the  same  values  for  the  circle  as  for  the  curve 

dx         dxl 

at  P. 


EXERCISES 

1.  The  length  of  arc  measured  from  a  fixed  point  on  a  certain  curve 
is  s  =  x2  +  x.     Find  the  slope  of  the  curve  at  x  =  2. 

2.  Can  x  =  cos  s,y  =  sin  s,  represent  a  curve  on  which  s  is  the  length 
of  arc  measured  from  a  fixed  point  ?  Can  x  =  sec  s,  y  =  tan  s,  represent 
such  a  curve? 

Find  the  radius  of  curvature  on  each  of  the  following  curves  at  the 
point  indicated: 

v3.   $  +  $=h    at  (0,6).  "  ■        ■-       * 


5.    r  =  ee,     at  6  = 


4.    x2  +  xy  +2/2  =  3,     at  (1,  1).      6.    r  =  a  (1  +  cos  6),     at  0  =  0. 

Find  an  expression  for  the  radius  of  curvature  at  any  point  of  each  of 
the  following  curves: 


(X  x\ 


9.   x  =  \  y2  -  \  In  y. 


8.   x  =  In  sec  y.  10.   r  —  a  sec2  \  9. 

11.   Show  that  the  radius  of  curvature  at  a  point  of  inflection  is 
infinite. 


Chap.  VII. 


GEOMETRICAL  APPLICATIONS 


77 


12.  A  point  on  the  circumference  of  a  circle  rolling  along  the  x-axis 
generates  the  cycloid 

x  =  a  (</>  —  sin  </>),         y  =  a  (1  —  cos  </>), 

a  being  the  radius  of  the  rolling  circle  and  </>  the  angle  through  which  it 
has  turned.  Show  that  the  radius  of  the  circle  of  curvature  is  bisected 
by  the  point  where  the  rolling  circle  touches  the  x-axis. 

13.  A  string  held  taut  is  unwound  from  a  fixed  circle.     The  end  of 
the  string  generates  a  curve  with  parametric  equations 

x  =  a  cos  0  +  ad  sin  0,         y  =  a  sin  0  —  ad  cos  0, 

a  being  the  radius  of  the  circle  and  0  the  angle  subtended  at  the  center 
by  the  arc  unwound.  Show  that  the  center  of  curvature  corresponding 
to  any  point  of  this  path  is  the  point  where  the  string  is  tangent  to  the 
circle. 

14.  Show  that  the  radius  of  curvature  at  any  point  (x,  y)  of  the  hypo- 

cycloid  x%  +  y3  =  a5  is  three  times  the  perpendicular  from  the  origin 
to  the  tangent  at  Or,  y). 

I  —  COS  X 

58.   Limit  of  —-It   is   shown   in   trigonometry 


that 


x 


1  —  cos  x  =  2  sin2 


Consequently, 


2.  „  x  /  .    x 

sin2- 
-  cos  x  2.x 

= =  sin  - 

x  x  2 


s,n- 


As  x  approaches  zero, 


.    x 
sin2 

X 

2 


x 
{    2    ) 


approaches  1.     Therefore 


,.     1  -  cos  x 

hm =  0  •  1  =  0. 

x=0  X 

59.  Derivatives  of  Arc  in  Polar  Coordinates.  —  The 
angle  from  the  outward  drawn  radius  to  the  tangent  drawn 
in  the  direction  of  increasing  s  is  usually  represented  by  the 
letter  ^. 


78 


DIFFERENTIAL  CALCULUS 


Chap.  VII. 


Let  r,  0  be  the  polar  coordinates  of  P,  and  r  +  Ar,  0  +  A0 
those  of  Q  (Fig.  59a).  Draw  QR  perpendicular  to  PR  and 
let  As  =  arc  PQ.     Then 

fj>i>n\      R®      (r+Ar)sinA0      ,    ,A.smA0    A0     As 
sin  (RPQ)  =  j~  = ^ =  (r+Ar) 


cos  (RPQ)  = 


PQ  PQ 

PR      (r  +  Ar)  cos  A0 

PQ~ 


A0       As    PQ 


PQ 


,AM  Ar      r  (1  —  cos  A0) 
=  cos  (A0)  ^ — ^ 


=  cos  (A0) 


Ar     As      r  (1  —  cos  A0)  A0      As 


As    PQ 


A0 


As     PQ 


Fig.  59a. 


Fig.  59b. 


As  A0  approaches  zero, 

r     rr>T>r\\      ,    v     sinA0          ..     l-cosA0  ..     As 

hm(#PQ)=^,  hm— ^-=1,  lim ^ =0,  hm^  =  l. 

The  above  equations  then  give  in  the  limit, 


sin  \p  = 


rdB 
ds 


dr 

cos  yp  =  -j-  ■ 
ds 


(59a) 


These  equations  show  that  dr  and  rdB  are  the  sides  of  a 
right  triangle  with  hypotenuse  ds  and  base  angle  \f/.  From 
this  triangle  all  the  equations  connecting  dr,  dd,  ds,  and  \J/ 
can  be  obtained.    The  most  important  of  these  are 


rdB 
tan  y  =  — r-  » 
dr 


ds2  =  dr2  +  r2  d02- 


(59b) 


Chap.  VII.  GEOMETRICAL  APPLICATIONS  79 

Example.     The  logarithmic  spiral  r  =  ae°. 
In  this  case,  dr  =  ae6  dd  and  so 

,  rdd 

tan  ^  =  -j-  =  1. 

The  angle  \f/  is  therefore  constant  and  equal  to  45  degrees. 
The  equation 

,      dr        1 

COS  \p  =    ,     = 


ds      V2 

dr  . 
shows  that  -=-  is  also  constant  and  so  r  and  s  increase  propor- 
tionally. 

EXERCISES 

Find  the  angle  \p  at  the  point  indicated  on  each  of  the  following  curves: 

1.  The  spiral  r  =  ad,  at  0  =  5  • 

o 

2.  The  circle  r  =  asin0at0=-:- 

4 

it 

3.  The  straight  line  r  =  a  sec  0,  at  0  =  ^  • 

4.  The  ellipse  r  (2  —  cos  0)  =  A;,  at  6  =  £• 

5.  The  lemniscate  r2  =  2  a2  cos  2  0,  at  6  =  f  71- . 

6.  Show  that  the  curves  r  =  ae0,  r  =  ae-0  are  perpendicular  at  each 
of  their  points  of  intersection. 

7.  Find  the  angles  at  which  the  curves  r  =  a  cos  d,  r  =  a  sin  2  6 
intersect. 

8.  Find  the  points  on  the  cardioid  r  =  a  (1  —  cos  8)  where  the  tan- 
gent is  parallel  to  the  initial  line. 

9.  Let  P  (r,  0)  be  a  point  on  the  hyperbola  r2  sin  2  0  =  c.  Show 
that  the  triangle  formed  by  the  radius  OP,  the  tangent  at  P,  and  the 
x-axis  is  isosceles. 

7T 

10.    Find  the  slope  of  the  curve  r  =  e2e  at  the  point  where  0  =  -y 

60.  Angle  between  Two  Directed  Lines  in  Space.  — 
A  directed  line  is  one  along  which  a  positive  direction  is 
assigned.     This  direction  is  usually  indicated  by  an  arrow. 


80 


DIFFERENTIAL  CALCULUS 


Chap.  VII. 


An  angle  between  two  directed  lines  is  one  along  the  sides 
of  which  the  arrows  point  away  from  the  vertex.  There  are 
two  such  angles  less  than  360  degrees,  their  sum  being  360 
degrees  (Fig.  60).     They  have  the  same  cosine. 

If  the  lines  do  not  intersect,  the  angle  between  them  is  de- 
fined as  that  between  intersecting  lines  respectively  parallel 
to  the  given  lines. 


y^x 


r0 


Fig.  60. 


Fig.  61. 


61.  Direction  Cosines.  —  It  is  shown  in  analytic  geome- 
try* that  the  angles  a,  0,  y  between  the  coordinate  axes  and 
the  line  PiP2  (directed  from  Pi  to  P2)  satisfy  the  equations 


cos  a  = 


x2  —  Xi 


cos/3  = 


2/2-  2/1 


cos  7  = 


Zi  —  Zi 


(61a) 


P,P2    '     ~ "        PiP2    '      ™  '        PiP2 

These  cosines  are  called  the  direction  cosines  of  the  line. 
They  satisfy  the  identity 

cos2  a  +  cos2  jS  +  cos2  7  =  1.  (61b) 

If  the  direction  cosines  of  two  lines  are  cos  ai,  cos  ft,  cos  71 
and  cos  a2,  cos  fa,  cos  72,  the  angle  6  between  the  lines  is  given 
by  the  equation 

cos  6  =  cos  ai  cos  a2  +  cos  j82  cos  &  +  cos  71  cos  72.  (61c) 
In  particular,  if  the  lines  are  perpendicular,  the  angle  6  is  90 
degrees  and 

0  =  cos  ai  cos  a2  -f  cos  jSi  cos  &  +  cos  71  cos  72.    (6 Id) 

*  Cf.  H.  B.  Phillips,  Analytic  Geometry,  Art.  64,  et  seq. 


Chap.  VII.  GEOMETRICAL  APPLICATIONS  81 

62.    Direction  of  the  Tangent  Line  to  a  Curve.  —  The 
tangent  line  at  a  point  P  of  a  curve  is  defined  as  the  limiting 

position    PT    approached    by  the    secant 
PQ  as  Q  approaches  P  along  the  curve. 

Let  s  be  the  arc  of  the  curve  measured 
from  some  fixed  point  and  cos  a,  cos  (3, 
cos  7  the  direction  cosines  of  the  tangent 
drawn  in  the  direction  of  increasing  s. 

If  .r,    y,   z  are   the   coordinates  of    P,  "G*  G2a> 

x  -f-  A x,  y  -\-  Ay,  z  -\-  Az}  those  of  Q,  the  direction  cosines  of 
PQ  are 

Ax       Ay        Az 

PQ'     PQ'     PQ' 

As  Q  approaches  P,  these  approach  the  direction  cosines  of 
the  tangent  at  P.     Hence 

,.      Ax      ,.     Ax  As 

COS  a  =  lim    t^:  =  lim  -r—vTiTS  • 

q=p  PQ  As\PQ 

On  the  curve,  x,  y,  z  are  functions  of  s.     Hence 

,.     Ax      dx        ..      As      ..       arc         .  * 

lim  -r—  =  -r  ,      lira  t^f:  =  hm  -, T  =  1. 

As       ds  PQ  chord 


Therefore 


Similarly, 


cos  a  =  -r  -  (62a) 


n,      dy  dz  foe.  N 

OMfl-gj,        cos7  =  ^.  (62a) 

These  equations  show  that  if  a  distance  ds  is  measured 
along  the  tangent,  dx,  dy,  dz  are  its  projections  on  the  coordi- 
nate axes  (Fig.  62b).     Since  the  square  on  the  diagonal  of  a 

*  The  proof  that  the  limit  of  arc/chord  is  1  was  given  in  Art.  53 
for  the  case  of  plane  curves  with  continuous  slope.  A  similar  proof 
can  be  given  for  any  curve,  plane  or  space,  that  is  continuous  in 
direction. 


82 


DIFFERENTIAL   CALCULUS 


Chap.  VII. 


rectangular  parallelopiped  is  equal  to  the  sum  of  the  squares 
of  its  three  edges, 

ds2  =  dx2  +  dy2  +  dz2.  (62b) 


Fig.  62b. 

Example.     Find  the  direction  cosines  of  the  tangent  to  the 
parabola 

x  =  at,        y  —  bt,        2  =  f  ct2 

at  the  point  where  t  =  2. 
At  t  =  2  the  differentials  are 

dx  =  a  dt,        dy  =  b  dt,        dz  =  \ctdt  =  c  dt, 

ds  =  ±  Vdx2  +  dy2  +  dz2  =  =b  Va2  +  b2  +  c2  ctt. 

There  are  two  algebraic  signs  depending  on  the  direction  s  is 
measured  along  the  curve.  If  we  take  the  positive  sign,  the 
direction  cosines  are 


63.   Equations  of  the  Tangent  Line.  —  It  is  shown  in 
analytic  geometry  that  the  equations  of  a  straight  line 


dx                 a 

dy                 b 

ds       Va2  +  b2  -f  c2  * 
dz                c 

ds       Va2  +  62  +  c2'                    I 

ds      Va2  +  b2  +  c2 

Chap.  VII.  GEOMETRICAL  APPLICATIONS  83 

through  a  point  Pi  (.1*1,  y\}  Z\)  with  direction  cosines  propor- 
tional to  A,  B,  C  are 

s-si      y-y\      z-zi  ,ao. 

-aT  =  —b-  =  -c-  (63) 

The  direction  cosines  of  the  tangent  line  are  proportional 
to  dx,  dy,  dz.  If  then  we  replace  A,  B,  C  by  numbers  pro- 
portional to  the  values  of  dx,  dy,  dz  at  Pi,  (63)  will  represent 
the  tangent  line  at  Pi. 

Example  1.   Find  the  equations  of  the  tangent  to  the  curve 

x  =  t,        y  =  t2,        z  =  iz 

at  the  point  where  t  =  1. 

The  point  of  tangency  is  t  =  1,  x±  =  1,  yi  =  1,  Zi  =  1. 

At  this  point  the  differentials  are 

dx  :dy  :dz  =  dt  :2tdt  :3t2  dt  =  1  :  2  :  3. 

The  equations  of  the  tangent  line  are  then 

x  —  1      y  —  1      z  —  1 
1  2      __ 3 

Ex.  2.  Find  the  angle  between  the  curve  3  x  +  2y  —  2  z 
=  3,  4:  x2  -\-  y2  =  2  z2  and  the  line  joining  the  origin  to 
(1,2,2). 

The  curve  and  line  intersect  at  (1,  2,  2).  Along  the  curve 
y  and  z  can  be  considered  functions  of  x.  The  differentials 
satisfy  the  equations 

3  dx  +  2  dy  —  2  dz  =  0,     8  x  dx  +  2  y  dy  =  4  z  dz. 

At  the  point  of  intersection  these  equations  become 

3  dx  +  2  cfy  -  2  dz  =  0,     8  dx  +  4  dy  =  8  efe. 

Solving  for  dx  and  d#  in  terms  of  dz,  we  get 

dx  —  2  dz,        dy  =  —2  dz. 

Consequently, 

ds  =  Vdx2  +  dy2  +  dz2  =  3  dz 
and 

dx      2  n      dy       —2  dz      1 

cosa  =  &=3'      C0S/3  =  dJ  =  ^-'      COS7  =  S  =  3- 


84  DIFFERENTIAL  CALCULUS  Chap.  VII. 

The  line  joining  the  origin  and  (1,  2,  2)  has  direction  cosines 
equal  to 

12       2 

3>     3»     3* 

The  angle  B  between  the  line  and  curve  satisfies  the  equation 

2-4  +  2 

cos  e  =  - — ^^  =  o. 

The  line  and  curve  intersect  at  right  angles. 

EXERCISES 

Find  the  equations  of  the  tangent  lines  to  the  following  carves  at  the 
points  indicated: 

IT 

1.  x  =  sec  t,     y  =  tan  t,    z  =  at,     at  t  =  -y 

2.  x  =  el,     y  =  e~l,     z  =  t2,     at  i  =  1. 

3.  x  =  el  sin  t,     y  =  el  cos  t,     z  =  kt,     at  t  =  -• 

4.  On  the  circle 

x  =  a  cos  0,     y  =  a  cos  (^  +  q7r)>     z  =  a  cos  (  0  +  ■=  ir  J 

show  that  ds  is  proportional  to  dd. 

5.  Find  the  angle  at  which  the  helix 

x  =  a  cos  0,     y  =  a  sin  0,     z  =  kd 

cuts  the  generators  of  the  cylinder  z2  +  y2  =  a?  on  which  it  lies. 

6.  Find  the  angle  at  which  the  conical  helix 

x  =  t  cos  t,     y  =  t  sin  t,     z  =  t 

cuts  the  generators  of  the  cone  x2  +  y2  =  z2  on  which  it  lies. 

7.  Find  the   angle  between   the   two   circles   cut  from  the  sphere 
x2  +  y2  +  z2  =  14  by  the  planes  x  —  y  +  z  =  0  and  x  -\-  y  —  z  =  2. 


CHAPTER  VIII 

VELOCITY   AND   ACCELERATION   IN   A   CURVED 

PATH 

64.  Speed  of  a  Particle.  —  When  a  particle  moves  along 
a  curve,  its  speed  is  the  rate  of  change  of  distance  along  the 
path. 

Let  a  particle  P  move  along  the  curve  A B,  Fig.  64.  Let  s 
be  the  arc  from  a  fixed  point  A  to  P.  The  speed  of  the  par- 
ticle is  then 

-s-  (64) 


Fig.  64. 


Fig.  65a. 


65.   Velocity  of  a  Particle.  —  The  velocity  of  a  particle 

at  the  point  P  in  its  path  is  defined  as  the  vector*  PT  tangent 

to  the  path  at  P,  drawn  in  the  direction  of  motion  with  length 

equal  to  the  speed  at  P.     To  specify  the  velocity  we  must 

then  give  the  speed  and  direction  of  motion. 

*  A  vector  is  a  quantity  having  Length  and  direction.    The  direction 
is  usually  indicated  by  an  arrow.     Two  vectors  arc  called  equal  when 

they  extend  along  the  same  line  or  along  parallel  lines  and   have  the 
same  length  and  direction. 

85 


86  DIFFERENTIAL  CALCULUS  Chap.  VIII. 

The  particle  can  be  considered  as  moving  instantaneously 
in  the  direction  of  the  tangent.  The  velocity  indicates  in 
magnitude  and  direction  the  distance  it  would  move  in  a 

unit  of  time  if  the  speed  and  direc- 
tion of  motion  did  not  change. 

Example.  A  wheel  4  ft.  in  di- 
ameter rotates  at  the  rate  of  500 
revolutions  per  minute.  Find  the 
speed  and  velocity  of  a  point  on  its 
rim. 

Let  OA  be  a  fixed  line  through  the 
center  of  the  wheel  and  s  the  distance  along  the  wheel  from 
OA  to  a  moving  point  P.     Then 

s  =  2  e  ft. 

The  speed  of  P  is 

^  =  2  %  =  2  (500)  2  7T  =  2000  t  ft./min. 
at         at 

Its  velocity  is  2000  x  ft./min.  in  the  direction  of  the  tangent 
at  P.  The  speeds  of  all  points  on  the  rim  are  the  same. 
Their  velocities  differ  in  direction. 

66.  Components  of  Velocity  in  a  Plane.  —  To  specify  a 
velocity  in  a  plane  it  is  customary  to  give  its  components, 
that  is,  its  projections  on  the  coordinate  axes. 

If  PT  is  the  velocity  at  P  (Fig.  66) ,  the  ^-component  is 

nn       d^  ds  ds  dx      dx 

PQ  =  PT  cos*  =  _COs*  =  ^  =  ^,      . 

and  the  ^/-component  is 

^m       T^m   •     ,       ds    .     ,      ds  dij      dv 

The  components  are  thus  the  rates  of  change  of  the  coordinates. 
Since 

PT2  =  PQ2  +  QT\ 


Chap.  VIII.      VELOCITY  AND  ACCELERATION 


87 


the  speed  is  expressed  in  terms  of  the  components  by  the 
equation 

(ds\2      (dx\2      (dy\2 

\dt)     :  \dt)  "^  \dt)  ' 


Fig.  66. 


Fig.  67. 


67.  Components  in  Space.  —  If  a  particle  is  moving 
along  a  space  curve,  the  projections  of  its  velocity  on  the 
three  coordinate  axes  are  called  components. 

Thus,  if  PT  (Fig.  67)  represents  the  velocity  of  a  point,  its 
components  are 


RT  =  PT  cos  t  = 


ds  dz      dz 
dt  ds      dt 


Since  PT2  =  PQ2  +  QR2  +  RT2}  the  speed  and  compo- 
nents are  connected  by  the  equation 


(day      hlxy      (dy\*      (dzY 
\dt)     '  \dt)  "*"  \dt)  "*"  \dt) 


88  DIFFERENTIAL  CALCULUS  Chap.  VIII. 

68.  Notation.  —  In  this  book  we  shall  indicate  a  vector 
with  given  components  by  placing  the  components  in  brack- 
ets. Thus  to  indicate  that  a  velocity  has  an  x-component 
equal  to  3  and  a  ^/-component  equal  to  —2,  we  shall  simply 
say  that  the  velocity  is  [3,  —2],  Similarly,  a  vector  in  space 
with  ^-component  a,  ^/-component  6,  and  2-component  c  will 
be  represented  by  the  symbol  [a,  b,  c]. 

Example  1.  Neglecting  the  resistance  of  the  air  a  bullet 
fired  with  a  velocity  of  1000  ft.  per  second  at  an  angle  of  30 
degrees  with  the  horizontal  plane  will  move  a  horizontal 
distance 

x  =  500  t  Vs 

and  a  vertical  distance 

y  =  500  t  -  16.1 12 

in  t  seconds.     Find  its  velocity  and  speed  at  the  end  of  10 
seconds. 
The  components  of  velocity  are 

^  =  500  V3,        ^  =  500  -  32.2  t 
at  at 

At  the  end  of  10  seconds  the  velocity  is  then 

V  =  1 500  V3,  178] 
and  the  speed  is 

jt  =  V(500  V3)2  +  (178)2  =  884  ft./sec. 

Ex.  2.  A  point  on  the  thread  of  a  screw  which  is  turned 
into  a  fixed  nut  describes  a  helix  with  equations 

x  —  r  cos  6,         y  =  r  sin  6,         z  =  kd, 

6  being  the  angle  through  which  the  screw  has  turned,  r  the 
radius,  and  k  the  pitch  of  the  screw.  Find  the  velocity  and 
speed  of  the  point. 


Chap.  VIII.       VELOCITY  AND  ACCELERATION 


89 


The  components  of  velocity  are 

dx  .      dd      dy  . dd      dz      .  dO 

-r.=  -r  sin  0  —  ,     -^  =  r  cos0tt,     -77  =  £-77. 
c/£  «£       dt  dt      dt         dt 

dd 
Since  -77  is  the  angular  velocity  co  with  which  the  screw  is 

rotating,  the  velocity  of  the  moving  point  is 

V  =  [—roj  sin  0,  ru  cos  0,  ku] 
and  its  speed  is 

ds 


dt 


=  Vr2ar  sin2  0  +  r2co2  cos2  0  +  &2co2  =  co  Vr2  +  A;2, 


which  is  constant. 


0 


Fig.  69a. 


Fig.  69b. 


69.    Composition   of  Velocities.  —  By   the  sum  of  two 

velocities  V\  and  V2  is  meant  the  velocity  Vi  +  V2  whose 
components  are  obtained  by  adding  corresponding  compo- 
nents of  Vi  and  V2.  Similarly,  the  difference  V2  —  V\  is  the 
velocity  whose  components  are  obtained  by  subtracting  the 
components  of  V\  from  the  corresponding  ones  of  V%. 
Thus,  if 

Vi  =  [«i,  61],         V2  =  [<h,  h2], 
Vi  +  F2  =  [«i  +  02, 61  +  62],       V2  -  Vi  =  [a2-  au  b2  -  fo]. 

If  V\  and  V2  extend  from  the  same  point  (Fig.  69a), 
V\  +  V2  is  one  diagonal  of  the  parallelogram  with  V\  and 
V2  as  adjacent  sides  and  V2  —  V\  is  the  other.  In  this  case 
V2  —  Vi  extends  from  the  end  of  V\  to  the  end  of  Fg. 


90 


DIFFERENTIAL  CALCULUS 


Chap.  VIII. 


By  the  product  mV  of  a  vector  by  a  number  is  meant  a 
vector  m  times  as  long  as  V  and  extending  in  the  same  direc- 
tion if  m  is  positive  but  the  opposite  direction  if  m  is  negative. 
It  is  evident  from  Fig.  69b  that  the  components  of  mV  are 
m  times  those  of  V. 

V  1 

The  quotient  —  can  be  considered  as  a  product  —  V.     Its 
m  m 

components  are  obtained  by  dividing  those  of  F  by  m. 

70.  Acceleration.  —  The  acceleration  of  a  particle  mov- 
ing along  a  curved  path  is  the  rate  of  change  of  its  velocity 

a       r      AF      dV 
A  =  Inn  -j—  =  — j-  - 
At=o  &t        at 

In  this  equation  AF  is  a  vector 

AF 
and  —r-r  is  obtained  by  divid- 
At  J 

ing   the    components    of   AT 

by  A*. 

Let  the  particle  move  from 

the  point  P  where  the  velocity 

is  V  to  an  adjacent  point  P' 

where  the  velocity  is  V  +  AF. 

dx    du 
The  components  of  velocity  will  change  from  -=r  ,  -~  to 


^+     di' 


Consequently, 
y  =  Ydx    dyl 

'  Idt '  dt]'      '    '  " '       L^ 
Subtraction  and  division  by  Af  give 


F  +  AF-g  +  Af 


dt~^      dt 


} 


AF 


-[ 


A^,  A^ 

6?£  <i£ 


} 


AF 
A* 


dx     A  dy 


dt 


dt 


At  '     A* 
As  A£  approaches  zero,  the  last  equation  approaches 

A 


dV      Vd?x    <Py~\ 
dt       Idt2'  dt2}' 


(70a) 


Chap.  VIII.      VELOCITY  AND  ACCELERATION 


91 


In  the  same  way  the  acceleration  of  a  particle  moving  in 
space  is  found  to  be 


.       \dH    d?y    d?z~\ 
Id?  '  df2 '  d#\ 


(701)) 


Equations  70a  and  70b  express  that  the  components  of  the 
acceleration  of  a  moving  particle  are  the  second  derivatives  of 
its  coordinates  with  respect  to 
the  time. 

Example-.  A  particle 
moves  with  a  constant 
speed  v  around  a  circle  of 
radius  r.  Find  its  velocity 
and  acceleration  at  each 
point  of  the  path. 

Let  6  =  AOP.  The  co- 
ordinates of  P  are 

x  =  r  cos  0, 
The  velocity  of  P  is 

V  =  I  —  r  si 


sm  0  jt  ,  r  cos  0 
dt 


Since  s  =  rd,  t;  =  v  —  r  -r± .     The  velocity  can  therefore  be 
dt  dt 

written 

V  =  [— vsinO,    v  cos0]. 

Since  v  is  constant,  the  acceleration  is 

dV 


A  = 


=  g(_„sin0),    !(f,COB*)] 


Replacing  -j-  by  -  ,  this  reduces  to 


v  n  v     '     fll       v 

=  \ cos  0, sin  6    =  - 

r  r  ]      r 

Now  [  —  cos0,   —  sin0]  is  a  vector  of  unit  length   directed 


A  =  \ cos0, sin  0    =  -  [  —  cos  0,  —sin  0]. 

r  r  r 


92  DIFFERENTIAL  CALCULUS  Chap.  VIII. 

along  PO  toward  the  center.     Hence  the  acceleration  of  P  is 
directed  toward  the  center  of  the  circle  and  has  a  magnitude 

equal  to  — . 

EXERCISES 

1.  A  point  P  moves  with  constant  speed  v  along  the  straight  line  y  =  a. 
Find  the  speed  with  which  the  line  joining  P  to  the  origin  rotates. 

2.  A  rod  of  length  a  slides  with  its  ends  in  the  x-  and  ?/-axes.  If  the 
end  in  the  a>axis  moves  with  constant  speed  v,  find  the  velocity  and  speed 
of  the  middle  point  of  the  rod. 

3.  A  wheel  of  radius  a  rotates  about  its  center  with  angular  speed  co 
while  the  center  moves  along  the  z-axis  with  velocity  v.  Find  the  velocity 
and  speed  of  a  point  on  the  perimeter  of  the  wheel. 

4.  Two  particles  Pi  (xi,  iji)  and  P2  (x2, y2)  move  in  such  a  way  that 

xx  =  1  +  2  t,  yi=2-3t*, 

x2  =  3  +  2P,  y2=     -  4  t\ 

Find  the  two  velocities  and  show  that  they  are  always  parallel. 

5.  Two  particles  Pi  (  Xi,  yi}  zx)  and  P2  (x2,  2/2,  22)  move  in  such  a  way 

that 

Xi  =  a  cos  6,yi  =  a  cos  (6  +  %  x),  Zi  =  a  cos  (6  +  f  it), 
x2  =  a  sin  9,y%  =  a  sin  (0  +  \  tt),  z2  =  a  sin  (d'-\-  f  k) . 

Find  the  two  velocities  and  show  that  they  are  always  at  right  angles. 

6.  A  man  can  row  3  miles  per  hour  and  walk  4.  He  wishes  to  cross 
a  river  and  arrive  at  a  point  6  miles  further  up  the  river.  If  the  river  is 
If  miles  wide  and  the  current  flows  2  miles  per  hour,  find  the  course  he 
shall  take  to  reach  his  destination  in  the  least  time. 

7.  Neglecting  the  resistance  of  the  air  a  projectile  fired  with  velocity 
[a,  b,  c]  moves  in  t  seconds  to  a  position 

x  =  at,     y  =  bt,     z  —  ct  —  |  gfi. 

Find  its  speed,  velocity,  and  acceleration. 

8.  A  particle  moves  along  the  parabola  x2  =  ay  in  such  a  way  that 

-r-  is  constant.    Show  that  its  acceleration  is  constant. 
at 

9.  When  a  wheel  rolls  along  a  straight  line,  a  point  on  its  circum- 
ference describes  a  cycloid  with  parametric  equations 

x  =  a  (<t>  —  sin  <£),        y  =  a  (1  —  cos  <j>), 

a  being  the  radius  of  the  wheel  and  <f>  the  angle  through  which  it  has 
rotated.     Find  the  speed,  velocity,  and  acceleration  of  the  moving  point. 


Chap.  VIII.  >  VELOCITY   AND   ACCELERATION  93 

10.  Find  the  acceleration  of  a  particle  moving  with  constant  speed  v 
along  the  cardioid  r  =  a  (1  —  cos  0). 

11.  If  a  string  is  held  taut  while  it  is  unwound  from  a  fixed  circle,  its 
end  describes  the  curve 

x  =  a  cos  6  -f-  a  9  sin  d,  y  =  a  sin  d  —  ad  cos  0, 

0  being  the  angle  subtended  at  the  center  by  the  arc  unwound.     Show 
that  the  end  moves  at  each  instant  with  the  same  velocity  it  would  have 

if  the  straight  part  of  the  string  rotated  with  angular  velocity  -t-  about 

the  point  where  it  meets  the  fixed  circle. 

12.  A  piece  of  mechanism  consists  of  a  rod  rotating  in  a  plane  with 
constant  angular  velocity  co  about  one  end  and  a  ring  sliding  along  the 
rod  with  constant  speed  v.  (1)  If  when  t  =  0  the  ring  is  at  the  center 
of  rotation,  find  its  position,  velocity,  and  acceleration  as  functions  of  the 
time.  (2)  Find  the  velocity  and  acceleration  immediately  after  t  =  tu 
if  at  that  instant  the  rod  ceases  to  rotate  but  the  ring  continues  sliding 
with  unchanged  speed  along  the  rod.  (3)  Find  the  velocity  and  acceler- 
ation immediately  after  t  =  h  if  at  that  instant  the  ring  ceases  sliding 
but  the  rod  continues  rotating.  (4)  How  are  the  three  velocities  re- 
lated?    How  are  the  three  accelerations  related? 

13.  Two  rods  AB,  BC  are  hinged  at  B  and  lie  in  a  plane.  A  is 
fixed,  AB  rotates  with  angular  speed  to  about  A,  and  BC  rotates  with 
angular  speed  2  a>  about  B.  (1)  If  when  t  =  0,  C  lies  on  AB  produced, 
find  the  path,  velocity,  and  acceleration  of  C.  (2)  Find  the  velocities 
and  accelerations  immediately  after  t  =  h  ii  at  that  instant  one  of  the 
rotations  ceases.  (3)  How  are  the  actual  velocity  and  acceleration 
related  to  these  partial  velocities  and  accelerations? 

14.  A  hoop  of  radius  a  rolls  with  angular  velocity  coi  along  a  horizon- 
tal line,  while  an  insect  crawls  along  the  rim  with  speed  ao>2.  If  when 
t  =  0  the  insect  is  at  the  bottom  of  the  hoop,  find  its  path,  velocity,  and 
acceleration.  The  motion  of  the  insect  results  from  three  simultaneous 
actions,  the  advance  of  the  center  of  the  hoop  with  speed  acoi,  the  rota- 
tion of  the  hoop  about  its  center  with  angular  speed  coi,  and  the  crawl 
of  the  insect  advancing  its  radius  with  angular  speed  o>2.  Find  the  three 
velocities  and  accelerations  which  result  if  at  the  time  t  =  h  two  of  these 
actions  cease,  the  third  continuing  unchanged.  How  are  the  actual 
velocity  and  acceleration  related  to  these  partial  velocities  and  accelera- 
tions? 


Tat^ 


CHAPTER  IX 

ROLLE'S  THEOREM  AND  INDETERMINATE 

FORMS 

71.  Rolle's  Theorem.  —  If  f  (x)  is  continuous,  there  is  at 
least  one  real  root  of  f  {x)  =  0  between  each  pair  of  real  roots 
off  (x)  =  0. 

To  show  this  consider 
the  curve 

V  =/(*)- 

Let  /  (x)  be  zero  at 
s  x  =  a  and  x  =  b.  Be- 
tween a  and  b  there 
must  be  one  or  more 
points  P  at  maximum 
distance  from  the  a>axis.      \t  such  a  point  the  tangent  is 

horizontal  and  so 

dy 


Fig.  71a. 


dx 


-  r  w  =  o. 


That  this  theorem  may  not  hold  if  f  (x)  is  discontinuous 
is  shown  in  Figs.  71b  and  71c.     In  both  cases  the  curve 


Fig.  71b. 


Fig.  71c. 


crosses  the  a>axis  at  a  and  b  but  there  is  no  intermediate 
point  where  the  slope  is  zero. 

94 


Chap.  IX.  ROLLE'S  THEOREM  95 

Example.     Show  that  the  equation 

x*  +  3x-6  =  0 

cannot  have  more  than  one  real  root. 
Let 

/  (x)  =  x3  +  3  x  —  6. 
Then 

f  (x)  =  3  a;2 +  3  =  3  (x2  +  1). 

Since  /'  (x)  does  not  vanish  for  any  real  value  of  x,  f  (x)  =  0 
cannot  have  more  than  one  real  root;  for  if  there  were  two 
there  would  be  a  root  of  /'  (x)  =  0  between  them. 

72.  Indeterminate  Forms.  —  The  expressions 

5,    -,    O.ao,     oo-oo,     l00,     0°,     oo° 
0      oo 

are  called  indeterminate  forms.  No  definite  values  can  be 
assigned  to  them. 

If  when  x  =  a  a  function  /  (x)  assumes  an  indeterminate 
form,  there  may  however  be  a  definite  limit 

lim/0). 

x==a 

In  such  cases  this  limit  is  usually  taken  as  the  value  of  the 
function  at  x  =  a. 
For  example,  when  x  =  0  the  function 

2x  =0 
x    ~  0* 

It  is  evident,  however,  that 

lim  —  =  lim  (2)  =  2. 

x=0     # 

This  example  shows  that  an  indeterminate  form  can  often  be 
made  definite  by  an  algebraic  change  of  form. 

73.  The  Forms  -  and  — .  — We  shall  now  show  that,  if 

0  oo 

f  (x) 

for  a  particular  value  of  the  variable  a  fraction  „  ,  x  assumes 

F(x) 

the  form  -  or  — ,  numerator  and  denominator  can  be  replaced 
0      oo  ^ 


96  DIFFERENTIAL  CALCULUS  Chap.  IX. 

by  their  derivatives  without  changing  the  value  of  the  limit 
approached  by  the  fraction  as  x  approaches  a. 

1.  Let  /'  (x)  and  F'  (x)  be  continuous  between  a  and  b. 
Iff  (°0  =  0,  F  (a)  =  0,  and  F  (b)  is  not  zero,  there  is  a  number 
X\  between  a  and  b  such  that 

f  (b)    r  fe) 


F  (6)      F'  (Xl) 
To  show  this  let  tMr  =  R.    Then 


(73a) 


F(b) 
f  (b)  -  RF  (b)  =  0. 

Consider  the  function 

/  (x)  -  RF  (x). 

This  function  vanishes  when  x  =  b.  Since  /  (a)  =  0, 
F  (a)  =  0,  it  also  vanishes  when  x  =  a.  By  Rolle's  Theorem 
there  is  then  a  value  Xi  between  a  and  b  such  that 

/'  fa)  -  RF'  (x,)  =  0. 

Consequently, 

/(&)  =  g  =  /'  fa) 

F(b)        F'(Xly 

which  was  to  be  proved. 

2.  Let  f  (x)  and  F'  (x)  be  continuous  near  a.    Iff  (a)  =  0 
and  F  (a)  =  0,  then 

52  £8 -22  £8-  (73b) 

For,  if  we  replace  6  by  #,  (73a)  becomes 

six)    r (xd 

F  (x)      F'  (xi) ' 

#i  being  between  a  and  x.  Since  Xi  approaches  a  as  #  ap- 
proaches a, 

hm  ^rK  =  hm  ,LT/ v,   \  =  hm  fjrf-f  • 

*-0  F  (X)         Xl=a  F'  (Xi)  x=a  F'  (x) 

3.  In  the  neighborhood  of  x  =  a,  let  /'  (#)  and  F'  (x)  be 


Chap.  IX.  ROLLE'S  THEOREM  97 

continuous  at  till  points  except  x  =  a.  If  f  (x)  and  F  (x) 
approach  infinity  as  x  approaches  a, 

Inn  Wj-^  =  Inn  J  ,  f  \  • 
xAa  F  (x)       x±a  F'  (x) 

To  show  this  let  c  be  near  a  and  on  the  same  side  as  x. 
Since  /  (x)  —  /  (c)  and  F  (x)  —  F  (c)  are  zero  when  x  =  c, 
by  Theorem  1, 

i  _  LVL 

f  (x.)       /(»)-/  (c)        /  (x)  /  (X) 

F'  (*0  "  F  (x)  -  F  (c)  ' "  F  (x)  F(c)' 

F(x) 

where  .Ti  is  between  x  and  c.  As  x  approaches  a,  /  (x)  and 
F  (x)  increase  indefinitely.  The  quantities  f(c)/f(x)  and 
F  (c)/F  (x)  approach  zero.  The  right  side  of  this  equation 
therefore  approaches 

x+a  F  (X) 

Since  .Ti  is  between  c  and  a,  by  taking  c  sufficiently  near  to 
a  the  left  side  of  the  equation  can  be  made  to  approach 

Since  the  two  sides  are  always  equal,  we  therefore  conclude 
that 

Inn  feV^  =  Inn  ^rK* 

x-0  F  (z)  x=a  F'  (X) 

sin  ^c 
Example   1.   Find  the  value  approached  by  as  a; 

x 

approaches  zero. 

Since  the  numerator  and  denominator  are  zero  when  x  =  0, 

we  can  apply  Theorem  2  and  so  get 

..     sin  x      ,.     cos  x      . 
Inn =  Inn — - —  =  1. 

i  =  0      X  x=0       J- 

I     — j—   PQg   T 

Ex.  2.    Find  the  value  of  lim  -, rz-  • 

«*,  (?r  -  x)2 


98  DIFFERENTIAL   CALCULUS  Chap.  IX. 

When  x  =  ir  the  numerator  and  denominator  are  both 
zero.     Hence 

,.     1  +  cosz       ..       (—  smx)        0 

lim  -, rr  =  hm  — —, r  =  -• 

x^    (tt  -  x)2        x=w  -2  (?r  -  x)      0 

Since  this  is  indeterminate  we  apply  the  method  a  second 
time  and  so  obtain 

,.         sin  z  ,.     cosz      1 

lim9? v>  =  lim  — 9  =  9  * 

x=v  2  [TT  —  X)  x=ir     —2  I 

The  value  required  is  therefore  \. 

\<\w  *\  x 
Ex.  3.   Find  the  value  approached  by  — as  x  ap- 
ian x 

IT 


proaches  ~  • 


7T 

When  x  approaches  ~  the  numerator  and  denominator  of 

this  fraction  approach  oo.     Therefore,  by  Theorem  3, 

lim  tan  3  £      , .     3  sec2  3  x      , .     3  cos2  x 

■k  -7 ■  =  hm— — = =  lim  — — — 

*=2    tan  x  sec2  x  cos2  3  x 

t  .  0 

When  x  is  replaced  by  -  the  last  expression  takes  the  form  -  • 

Therefore 

3  cos2  x      ..         6  cos  #  sin  z 
hm  — tttz—  =  hm 


cos2  3  a;  6  cos  3  x  sin  3  x 

cos2  x  —  sin2  x  1 


=  lim 


3  (cos2  3  x  -  sin2  3  x)  ' "  3 


74.   The  Forms  0  •  oo  and  oo  —  oo.  •—  By  transforming 

the  expression  to  a  fraction  it  will  take  the  form  ^  or  — 

For  example, 

sins 


Chap.  IX.  ROLLE'S  THEOREM  99 

has    the    form    0  •  oo    when   x  =  0.      It    can,  however,   be 
written 

.  \nx 

x  In  x  =  — —  i 

x 

which  has  the  form  —  • 

00 

The  expression 

sec  x  —  tan  x 

IT 

has  the  form  oo  —  oo  when  x  =  -•      It  can,  however,  be 

written 

1         sin  x      1  —  sin  x 


sec  x  —  tan  x  = 


cos  x      cos  x  cos  x 


which  becomes  ^  when  x  —  ~  • 

75.  The  Forms  0°,  1"°,  oo°.  —  The  logarithm  of  the  given 
function  has  the  form  0  •  oo.  From  the  limit  of  the  log- 
arithm the  limit  of  the  function  can  be  determined. 

i 

Example.     Find  the  limit  of   (1  +  x)x   as  x  approaches 

zero. 

i 

Let  y  =  (1  +  xf. 


Then 


i  *  i    n    i     \      In  (1  +  x) 

\ny  =  -ln(l+s)  =  -  — — 

X  Jj 


When  z  is  zero  this  last  expression  becomes  ^  .     Therefore 

v     ln(l-f-:r)       ..         1  1 

lim    — =  lim  —  t—  =  1. 

x=0  X  1   T"  X 

The  limit  of  In  y  being  1,  the  limit  of  y  is  e/ 


100 


DIFFERENTIAL  CALCULUS 


Chap.  IX. 


EXERCISES 

1.   Show  by  Rolle's  Theorem  that  the  equation 

z4-4x-l=0 

cannot  have  more  than  two  real  roots. 

Determine  the  values  of  the  following  limits: 


x9  -  1 


4. 

x 1?  *10  -  i 

3. 

x11  —  1 

Lim =-• 

x=l    x  —  1 

4. 

T  .      1  —  COS  X 

Lim                    • 

x±o      sin  x 

5. 

T  .     ex  -  ea 

Lim 

x==a   x        a 

6. 

T  .     tan  x  —  x 

j-jiiii            .       • 

x=o  x  —  sm  x 

7. 

X2  cos  X 

IjIIII                           -  • 

Xd=0  cos  x  -  1 

8. 

Limln(:C-2) 

IjIIII                       n 

x=3       X  -  3 

9. 

In  cos  x 

Lim 

x=0        x 

n 

T  .       sin2  -kx 

I  ,i  m • 

M  (X  -  2)2 


11.    Lim 


1  -f  cos  x  —  sin  x 


x=  — 

2 


7T  cos  x  (2  sin  x  —  1) 
logio  (sin  x  —  sin  a) 


12.    Lim  . 

x=a  logio  (tan  x  —  tan  a) 


13.    Lim 

x=0 


6  sin  x  —  6  x  +  re3 


zJ 


., .     T  .     sec2  d>  —  2  tan  </> 

14.  Lim  — t-2- — - 

v      1  +  cos  4  0 

**4 

15.  Lim — : — • 

ii0  cot  X 

16.  Lim • 

£  =  00       X 


.  _    T  .     sec  3  x  —  x 
17.    Lim 


x= 


18.    Lim 


1  +  tan  2  x 


*!■»(■ +i) 


19.  Lim  x  cot  x. 

x=0 

20.  Lim  tan  a:  cos  3  #. 

X=2 

21.  Lim  (;r  +  a)ln(l+-Y 

X=oo  \  #/ 

22.  Lim  (z  —  3)  cot  (tx). 

x=3 


24. 

Lim  .r^2 . 

x=0 

25. 

x±0\x       ex  -  1/ 

26. 

Lim  (cot  x  —  In  x) . 
x=0 

27. 

JL/lIIi   1    IcUl.t            .                     . 

»  L             sin  x—  sin2  x 

X=2 

28. 

Lim  r*. 

x=0 

29. 

Lim(sinx)tanx. 

30. 

.  if 

X=2 

1 

Lim  (1  +  ax)x. 
x=0 

1 

] 


31.   Lim  (xm  -  am)  in  *. 

s=oo 


»   - 

I  i 


CHAPTER  X 

SERIES   AND    APPROXIMATIONS 

76.  Mean  Value  Theorem.  —  ///  (x)  and  /'  (x)  are  con- 
tinuous from  x  =  a  to  x  =  b,  there  is  a  value  X\  between  a  and  b 
such  that 

/(&)-/  (a) 


=  /'  (*o. 


(76) 


b  —  a 

To  show  this  consider  the  curve  y  =  f  (x).     Since  f  (a) 
and  /  (6)  are  the  ordinates  at  x  =  a  and  x  =  b, 

/(«•)-/  (a) 


b  —  a 


=  slope  of  chord  AB. 


On  the  arc  .4  B  let  Pi  be  a  point  at  maximum  distance  from 


Fig.  76. 

the  chord.     The  tangent  at  Pi  will  be  parallel  to  the  chord 
and  so  its  slope/'  (zi)  will  equal  that  of  the  chord.  •   Therefore 

/  (b)  -  f  (a) 


b  —  a 


=  /'  (ft), 


which  was  to  be  proved. 

Replacing  b  by  x  and  solving  for/  (x),  equation  (76)  becomes 

f(x)  =f(a)  +  (x-a)f'(xl), 
101 


102  DIFFERENTIAL   CALCULUS  Chap.  X. 

Xi  being  between  a  and  a;.  -  This  is  a  special  case  of  a  more 
general  theorem  which  we  shall  now  prove. 

77.  Taylor's  Theorem.  —  If  f  (x)  and  all  its  derivatives 
used  are  continuous  from  a  to  x,  there  is  a  value  X\  between  a  and  x 
such  that 

f  (x)  -/(a)  +  (x  -  a)/'  (a)  +  %LZ°£.y  (a) 

+  (^fi-  /'"  (a)  +  ■  ■  ■  +^-Z^>lfn  (Xl). 
To  prove  this  let 
*(x)=/(x)-/(a)-(x-a)/'(a) 

- ^r^  /" («) V-i)!*  /"~I  (a)' 

It  is  easily  seen  that 

</>  (a)  =  0,     0'  (a)  =  0,     0"  (a)  =  0, 

.  .  .  <f>n~l  (a)  =  0,     cf>n  (x)  =  fn  (x). 

When  x  =  a  the  function 

»(g) 
(x  —  a)n 

therefore  assumes  the  form  -.     By  Art.  73  there  is  then  a 

0         J 

value  Z\  between  a  and  x  such  that 

^  O)  0'  (Zi) 


(x  —  a)n      n  (zi  —  a)n~l 
This  new  expression  becomes  -  when  z\  =  a.     There  is  con- 
sequently a  value  z2  between  zx  and  a  (and  so  between  x  and  a) 
such  that 

»'(*)         =  *"  (*) 

n  (21  —  a)n_1      7i  (w  —  1)  (%  —  a)n~2 

A  continuation  of  this  argument  gives  finally 

4>{x)      =  <t>n  (Zn)  =  fn  (zn) 
(x  —  a)n  n\  n\ 


Chap.  X.  SERIES  AND  APPROXIMATIONS  103 

zn  being  between  x  and  a.     If  Xi  =  zn  we  then  have 

Equating  this  to  the  original  value  of  <t>  (x)  and  solving  for 
/(a;),  we  get 

/(»)=/(a)  +  (*-o)/'(o) 

+  ^-^-V'(a)+  •  •  •'  +  ^r->(*i). 

which  was  to  be  proved. 
Example.     Prove 

mx      {x      i)  2^3  4^4 

where  Zi  is  between  1  and  x. 
When  x  =  1  the  values  of  In  x  and  its  derivatives  are 

f(x)=ln(x),  /(1)  =  0, 

/'  (*)  =  i.  /'  (1)  =  1, 

/'"(*)  =  |.  /'"(1)=2, 

/""(*)  =  -§.  /""(*>  =  "^ 

Taking  a  =  1,  Taylor's  Theorem  gives 
Ins  =  0  +  1  (*-  1)  -|(x-  l)*  +  g(*-  1)3  _  A  &L=_1>_\ 

which  is  the  result  required. 

78.   Approximate  Values  of  Functions.  —  The  last  term 
in  Taylor's  formula 

^T^  /"  (x.)    =    ft. 


104  DIFFERENTIAL  CALCULUS  Chap.  X. 

is  called  the  remainder.     If  this  is  small,  an  approximate 
value  of  the  function  is 

f(x)=f(a)  +  (x-a)f'(a) 

+  ^y^  /"  («)+•••+  %\%  f"~l  (a)' 

the  error  in  the  approximation  being  equal  to  the  remainder. 

To  compute/  (x)  by  this  formula,  we  must  know  the  values 
of/  (a),  /'  (a),  etc.  We  must  then  assign  a  value  to  a  such  that 
f  (a)>  f  (fl)>  etc-i  are  known.  Furthermore,  a  should  be  as 
close  as  possible  to  the  value  x  at  which  f  (x)  is  wanted.  For, 
the  smaller  x  —  a,  the  fewer  terms  (x  —  a)2,  (x  —  a)3,  etc., 
need  be  computed  to  give  a  required  approximation. 

Example  1.    Find  tan  46°  to  four  decimals. 

The  value  closest  to  46°  for  which  tan  x  and  its  derivatives 

are  known  is  45°.     Therefore  we  let  a  =  -r- 

4 

f(x)  =  tana;,  ^(l)=  lf 

/' (*)  =  sec2  x,  f'(S)=2' 

/"  (x)  =  2  sec2  x  tan  x,  /"  (£\  =  4, 

J'"  (x)  =  2  sec4  x  +  4  sec2  x  tan2  x. 
Using  these  values  in  Taylor's  formula,  we  get 


7T1 

X  —  -M 


and 

ta„46°  =  l+2(I|j)  +  2(I|j)2=1.0355 

approximately.      Since  Xi  is  between  45°  and  46°,  /'"  (#i) 
does  not  differ  much  from  j 

/'"(45°)  =8  +  8  =  16. 


Chap.  X.  SERIES   AND   APPROXIMATIONS  105 

The  error  in  the  above  approximation  is  thus  very  nearly 
16  /  w  \3         8  1 

6  \is6J  <  sjwy  <  4^000  "  0-000025- 

It  is  therefore  correct  to  4  decimals. 

Ex.  2.    Find  the  value  of  e  to  four  decimals. 

The  only  value  of  x  for  which  e?  and  its  derivatives  are 
known  is  x  =  0.     We  therefore  let  a  be  zero. 

/  0)  =  e*,    /'  (x)  =  e*,    f"  (x)  =  e, ,   f»  (x)  =  #, 

/(0)  -  1,     f  (0)  =  1,     /"  (0)  =  1, ,  f*  (Xl)  =  e*>. 

By  Taylor's  Theorem, 

_       -    .        .  x2      x3  xn~l       .  xnex* 

e*=l  +  x  +  -  +  -  +  •  •  ■  + —  H r  • 

2!      3!  (n  —  1)1        n\ 

Letting  x  =  1,  this  becomes 

11  1  e*i 

i-hit-2!  -r-3!  -r-  +(»_i)|+n! 

In  particular,  if  n  =  2, 

e  =  2  +  }  £**. 
Since  Zi  is  between  0  and  1.  e  is  then  between  2|  and  2  + 
J  e,  and  therefore  between  2|  and  4.      To  get  a  better  ap- 
proximation let  n  =  9.     Then 

e  =  1  +  1  +  1-  +  ^+  ...  +1=2.7183 

approximately,  the  error  being 

px*       p        A- 

9!  '  9!  =  9!  <  -O0002- 
The  value  2.7183  is  therefore  correct  to  four  decimals. 

EXERCISES 

Determine  the  values  of  the  following  functions   correct  to  four 
decimals: 

1.  sin  5°.  5.    sec  (10°). 

2.  cos  32°.  6.    In  (&). 

3.  cot  43°.  7.    y/e. 

4.  tan  58°.  8.    tan"1  (&). 
9.   Given  In  3  =  1.0986,  In  5  =  1.6094,  find  In  17. 


106  DIFFERENTIAL  CALCULUS  Chap.  X. 

79.   Taylor's  and  Maclaurin's  Series.  —  As  n  increases 
indefinitely,  the  remainder  in  Taylor's  formula 

Rn  =  fr  -  a>>  {Xl)  ' 

n 
often  approaches  zero.     In  that  case 

/ (z)  =  lim  \f  (a)  +  (x-a)  f(a)+  ■  ■  ■  +  (f~1w V""'  (a)T 

This  is  usually  written 

/(*)-/  (a)  +  (x  -  a)/'  (a)  +  ^=^-2 /"  (a) 

+^ir-V"(a)+  ••  -, 

the  dots  at  the  end  signifying  the  limit  of  the  sum  as  the 
number  of  terms  is  indefinitely  increased.  Such  an  infinite 
sum  is  called  an  infinite  series.  This  one  is  called  Taylor's 
Series. 

In  particular,  if  a  =  0,  Taylor's  Series  becomes 

/(x)=/(0)+x/'(0)+|/"(0)+^/"'(0)+  .... 

This  is  called  Maclaurin's  Series. 

Example.     Show  that  cos  x  is  represented  by  the  series 

/y»a  /y»4  /y»0 

COSX=1-2!+4!-6!+  •'•• 

The  series  given  contains  powers  of  x.     This  happens  when 
a  =  0,  that  is,  when  Taylor's  Series  reduces  to  Maclaurin's. 

/  (x)  =  cos  x,  f  (0)  =  1, 

f(x)  =  -sins,  /'  (0)=0, 

/"(a)  =  -cosx,  /"(0)  =  -1, 

/'"  (a?)  =  sin  x,  /"'  (0)  =  0, 

/""  (x)  =  cosz,  /""  (0)  =  1. 

These  values  give 

/v»2  /v*4  /y»7t 

cos  x  =  1  -  2|  +  j]  -   •  •  •  ±  -,  /n  (*i). 


Chap.  X.  SERIES  AND  APPROXIMATIONS  107 

The  nth  derivative  of  cos  x  is  icos  x  or  ±sin  x,  depending 
on  whether  n  is  even  or  odd.  Since  sin  x  and  cos  x  are  never 
greater  than  l,/n  (xi)  is  not  greater  than  1.     Furthermore 

Xn        X    X    X  x 

rH  ~  T'2*3*   '  *  "n 
can  be  made  as  small  as  you  please  by  taking  n  sufficiently 
large.     Hence  the  remainder  approaches  zero  and  so 

-7*2  /y»4  sy*0 

cosx=l-2T  +  i]-gI+  •••, 
which  was  to  be  proved. 

EXERCISES 

.  X3    .     Xs         X7     . 

1.  gmx  =  x--+-rTl  +  .... 

o  !  x   f        A        l     (        A\      X     /         *Y_i 

3.  2*  =  l+xln2+^  +  M^+.... 

^vi2  -j»3  /y*5  /v»6  Q  j^7 

4.  ^sina^  *  +  2  .  ^  +  2  ^  -  4^-8  --  —  +.... 

_      .                 1(            2x*      4x*      4:x*  .  8x7   , 
6.   e*  cos  *  =  1  +  z  -  ^-  -  -^ 5T  +  -yr  + 

6.  (a  +  x)n  =  a"  +  nan~lx  +  n  ^n~  ^  a^x2  +  •  •  •  ,  if  M*<  |a|. 

7.  VS-2+^-^^«^-...,if|*-4|<l. 

8.  lnx  =  ln3+^-i^>^)!_...,if,,-3|<l. 

9.  ln(:r  +  5)  =  In  6 +^-^=^+^^ ,if|x-l|<l. 

80.  Convergence  and  Divergence  of  Series. —  An  in- 
finite series  is  said  to  converge  if  the  sum  of  the  first  n  terms 
approaches  a  limit  as  n  increases  indefinitely.  If  this  sum 
does  not  approach  a  limit,  the  series  is  said  to  diverge.    * 

The  series  for  sin  x  and  cos  x  converge  for  all  vaiues  of  x. 
The  geometrical  series 

a  +  ar  +  ar2  +  ar3  +  cuA  +   •  •  • 

*  The  symbol  |.r|  is  used  to  represent  the  numerical  value  of  x  with- 
out its  algebraic  sign.     Thus,  |  —  3  |  =  |  3  |  =  3. 


108  DIFFERENTIAL  CALCULUS  Chap.  X 

converges  when  r  is  numerically  less  than  Y.    For  the  sum  of 
the  first  n  terms  is 

Sn  —  a  +  ar  +  ar2  +  •  •  ■  -J-  arn~l  =  a 

1  —  r 

If  r  is  numerically  less  than  1,  rn  approaches  zero  and  Sn 
approaches 

1  —  r 

as  n  increases  indefinitely. 
The  series 

1-1  +  1-1  +  1-1+..  • 

is  divergent,  for  the  sum  oscillates  between  0  and  1  and  does 
not  approach  a  limit.     The  geometrical  series 

1  +  2  +  4  +  8  +  16+  •  •  • 

diverges  because  the  sum  increases  indefinitely  and  so  does 
not  approach  a  limit. 

81.  Tests  for  Convergence.  —  The  convergence  of  a 
series  can  often  be  determined  from  the  problem  in  which  it 
occurs.     Thus  the  series 

/v»2  /y»4  /y»D 

2!  +  4!  ~  6!  + 

converges  because  the  sum  of  n  terms  approaches  cos  x  as  n 
increases  indefinitely. 

The  terms  near  the  beginning  of  a  series  (if  they  are  all 
finite)  have  no  influence  on  the  convergence  or  divergence  of 
the  series.  This  is  determined  by  terms  indefinitely  far  out 
in  the  series. 

82.  General  Test.  —  For  the  series 

Ui  +  u2  +  u3  +   •  •   •   +  un  +   •  •  • 

to  converge  it  is  necessary  and  sufficient  that  the  sum  of  terms 
beyond  un  approach  zero  as  n  increases  indefinitely. 

For,  if  the  series  converges,  the  sum  of  n  terms  must  ap- 
proach a  limit  and  so  the  sum  of  terms  beyond  the  nth  must 
approach  zero. 


y 


Chap.  X.  SERIES    AM)    APPROXIMATIONS  109 

83.  Comparison  Test.  —  A  series  is  convergent  if  beyond 
a  certain  point  its  terms  are  in  numerical  value  respectively  less 
than  those  of  a  convergent  scries  whose  terms  are  all  positive. 

For,  if  a  series  converges,  the  sum  of  terms  beyond  the  nth 
will  approach  zero  as  n  increases  indefinitely.  If  then  another 
series  has  lesser  corresponding  terms,  their  sum  will  approach 
zero  and  the  series  will  converge. 

84.  Ratio  Test.  —  If  the  ratio  — —  of  consecutive  terms 

un 

approaches  a  limit  r  as  n  increases  indefinitely,  the  series 

-     Ill  +  Ui  +  U3  +    •    •    •    +  Un  +  Un-H  +     '    '    ' 

is  convergent  if  r  is  numerically  less  than  1  and  divergent  if  r  is 
numerically  greater  than  1. 

Since  the  limit  is  r,  by  taking  n  sufficiently  large  the  ratio 
of  consecutive  terms  can  be  made  as  nearly  r  as  we  please. 
If  r  <  1 ,  let  r\  be  a  fixed  number  between  r  and  1 .  We  can 
take  n  so  large  that  the  ratio  of  consecutive  terms  is  less  than 
n.    Then 

Un+1   <  nUn,    Un+2   <  Wn+1   <  r^Un,    etc. 

Beyond  un  the  terms  of  the  given  series  are  therefore  less  than 
those  of  the  geometrical  progression 

Un  +  TlUn  +  n2Un  +     •    •    • 

which  converges  since  i\  is  numerically  less  than  1.  Con- 
sequently the  given  series  converges. 

If,  however,  r  is  greater  than  1,  the  terms  of  the  series  must 
ultimately  increase.  The  terms  do  not  then  approach  zero 
and  their  sum  cannot  approach  a  limit. 

Example.     Find  for  what  values  of  x  the  series 

x  +  2x2  +  3x3  +  4x4  +  •  •  • 
converges. 

The  ratio  of  consecutive  terms  is 


Un+i  =  (n  +  1)  xn+l      / 
un  nxn  \         n 


-)  x. 


110  DIFFERENTIAL  CALCULUS  Chap.  X. 

The  limit  of  this  ratio  is 

r  =  lim  ( 1  H — )  x  =  x. 

The  series  will  converge  if  x  is  numerically  less  than  1. 

85.   Power  Series.  —  A  series  of   powers  of  (x  —  a)  of 
the  form 

P  (x)  =  do  +  a\  (x  —  a)  +  (h  (x  —  a)2  +  a3  (x  —  aY  +  •  •  •  , 

where  a,  a0,  ai,  02,  etc.,  are  constants,  is  called  a  "power  series. 
If  a  power  series  converges  when  x  =  b,  it  will  converge  for 
all  values  of  x  nearer  to  a  than  b  is,  that  is,  such  that 

\x  —  a\  <  \b  —  a\. 

In  fact,  if  the  series  converges  when  x  =  b,  each  term  of 

a0  +  ai  (6  —  a)  +  a2  (b  —  a)2  +  a3  (6  —  a)3  +  •  •  • 

will  be  less  than  a  maximum  value  M ,  that  is, 

\an  (b  -  a)n\  <M. 
Consequently, 

The  terms  of  the  series 

«o  +  0*1  (%  —  a)  +  (h  (x  —  a)2  +  a3  (x  —  a)3  +  •  •  • 
are  then  respectively  less  than  those  of  the  geometrical  series 

M+  iT \\x-a\+ \T r2l:r_al2+T^ d^-al3+  '  '  ' 

\b  —  a\ '  \b  —  a\2  '  [6  —  aj3 

in  which  the  ratio  is 

\x  —  a 


\b  -  a\ 

If  then  \x  —  a\  <\b  —  a\,  the  progression  and  consequently 
the  given  series  will  converge. 

If  a  power  series  diverges  when  x  =  b,  it  will  diverge  for  all 
values  of  x  further  from  a  than  b  is,  that  is,  such  that 

\x  —  a\  >  \b  —  w|.     . 


Chap.  X.  SERIES   AND   APPROXIMATIONS  111 

For  it  could  not  converge  beyond  6,  since  by  the  proof  just 
given  it  would  then  converge  at  6. 

This  theorem  shows  in  certain  cases  why  a  Taylor's  Series 
is  not  convergent.     Take,  for  example,  the  series 

/>*2  /y*3  /v4 

\n(i  +  X)  =  x--+---^+ .... 

As  x  approaches  —1,  In  (1  -\-  x)  approaches  infinity.  Since 
a  convergent  series  cannot  have  an  infinite  value,  we  should 
expect  the  series  to  diverge  when  x  =  —I.  It  must  then 
diverge  when  z  is  at  a  distance  greater  than  1  from  a  =  0. 
The  series  in  fact  converges  between  x  =  —1  and  x  =  1  and 
diverges  for  values  of  x  numerically  greater  than  1. 

86.  Operations  with  Power  Series.  —  It  is  shown  in 
more  advanced  treatises  that  convergent  series  can  be  added, 
subtracted,  multiplied  and  divided  like  polynomials.  In 
case  of  division,  however,  the  resulting  series  will  not  usually 
converge  beyond  a  point  where  the  denominator  is  zero. 

Example.     Express  tan  x  as  a  series  in  powers  of  x. 

We  could  use  Maclaurin's  series  with  /  (x)  =  tan  x.  It  is 
easier,  however,  to  expand  sin  x  and  cos  x  and  divide  the  one 
by  the  other  to  get  tan  x.     Thus 

x3  ,    x5 
x  -  77  + 


sinrc  6   '   120  .  x*  .  2  x5  . 

tanz= = s =x+^  -T--TF--T- 

cos  x       1       ^  _    x  3       15j 

2+2i~ 


EXERCISES 


1.   Show  that 


ln(11-^|)  =  ln(l+x)-ln(l-x)=2^+f  +f+f+  "  •  )i 

and  that  the  series  converges  when  \x\  <  1. 
2.    By  expanding  cos  2  x,  show  that 

.              1  —  cos  2  x      nx2          xA           Xs 
81n  x  = 2 '  =  2  2!  "  2  4!  +  2   6! # 

Prove  that  the  series  converges  for  all  values  of  x. 


112  DIFFERENTIAL  CALCULUS  Chap.  X. 

3.  Show  that 

(1  +e*)2  =  l+2ex-J-e2a:  =  4  +  4x  +  3x2+|x3+|x4+*-- 

and  that  the  series  converges  for  all  values  of  x. 

4.  Given  /  (x)  =  sin-1  x,  show  that 

f  (x)  =      J_  =  (1  -  *)-*. 
vl  -  X2 

Expand  this  by  the  binomial  theorem  and  determine  f"  (x),  etc.,  by 
differentiating  the  result.     Hence  show  that 

.  _,              ,   1  x*  .   1    3  x5  .  1    3   5  x7  . 
sin  Y  x  =  x  -\ —  — —  •-  — u_.-. u  .  .  . 

t23t245t2    467f 

and  that  the  series  converges  when  \x\  <  1. 

5.  By  a  method  similar  to  that  used  in  Ex.  4,  show  that 

/y*0  spj  rpl 

tan"1  z  =  z--2  +  --y+... 

and  that  the  series  converges  when  \x\  <  1. 

6.  Prove 

sec  x  =  =  1  +  -  +  --.  x*  +  •  •  •  . 

cos  x  2       24 

For  what  values  of  x  do  you  think  the  series  converges? 


CHAPTER   XI 

PARTIAL   DIFFERENTIATION 

87.  Functions  of  Two  or  More  Variables.  —  A  quantity 
u  is  called  a  function  of  two  independent  variables  x  and  y, 

u  =/(x,  y), 

if  u  is  determined  when  arbitrary  values  (or  values  arbitrary 
within  certain  limits)  are  assigned  to  x  and  y. 
For  example, 

u  =  Vl  —  x2  —  y2 

is  a  function  of  x  and  y.  If  u  is  to  be  real,  x  and  y  must  be  so 
chosen  that  x2  +  y2  is  not  greater  than  1.  Within  that 
limit,  however,  x  and  y  can  be  chosen  independently  and  a 
value  of  u  will  then  be  determined. 

In  a  similar  way  we  define  a  function  of  three  or  more  in- 
dependent variables.  An  illustration  of  a  function  of  vari- 
ables that  are  not  independent  is  furnished  by  the  area  of  a 
triangle.  It  is  a  function  of  the  sides  a,  6,  c  and  angles  A,  B, 
C  of  the  triangle,  but  is  not  a  function  of  these  six  quantities 
considered  as  independent  variables;  for,  if  values  not  be- 
longing to  the  same  triangle  are  given  to  them,  no  triangle 
and  consequently  no  area  will  be  determined. 

The  increment  of  a  function  of  several  variables  is  its  in- 
crease when  all  the  variables  change.     Thus,  if 

u  =  f(x,  y), 

u  +  Aw  =  /  (x  +  Az,  y  +  Ay) 

and  so 

Aw  =  /  (x  +  Ax,  y  +  A?/)  -  /  (x,  y). 

A  function  is  called  continuous  if  its  increment  approaches 
zero  when  all  the  increments  of  the  variables  approach  zero. 

113 


114  DIFFERENTIAL   CALCULUS  Chap.  XI. 

88.  Partial  Derivatives.  —  Let 

.  u  =  /  fo  y) 

be  a  function  of  two  independent  variables  x  and  y.  If  we 
keep  y  constant,  u  is  a  function  of  x.  The  derivative  of  this 
function  with  respect  to  x  is  called  the  partial  derivative  of  u 
with  respect  to  x  and  is  denoted  by 

fx     or   fAx,y). 

Similarly,  if  we  differentiate  with  respect  to  y  with  x  con- 
stant, we  get  the  partial  derivative  with  respect  to  y  denoted 

by 

—     or    fy(x,y). 

For  example,  if 

u  =  x2  +  xy  —  y2, 

then 

^  =  2x  +  y,     Yy  =  x~2y. 

Likewise,  if  u  is  a  function  of  any  number  of  independent 
variables,  the  partial  derivative  with  respect  to  one  of  them 
is  obtained  by  differentiating  with  the  others  constant. 

89.  Higher  Derivatives.  —  The  first  partial  derivatives 

are   functions   of   the   variables.     By   differentiating   these 

functions  partially,  we  get  higher  partial  derivatives. 

du 
For  example,  the  derivatives  of  —  with  respect  to  x  and  y 

ox 


are 


d    /du\       d2u       d    fdu\        d2u 


fdu\  = 
,dx) 


Similarly, 


du\         d2u  d    /du\       d2u 


dx  \dx/       dx2      dy  \dxj      dy  dx 

_d    /du\  = 
dx  \dyj      dx  dy '     dy  \dy)       dy2 

It  can  be  shown  that 

d2u  d2u 

dx  dy      dy  dx ' 


Chap.  XI.  PARTIAL   DIFFERENTIATION  115 

if  both  derivatives  are  continuous,  that  is,  partial  derivatives 
are  independent  of  the  order  in  which  the  differentiations  are 
performed* 

Example,     u  =  x2y  +  xy%. 

-  =  2  .n,  +  y>,      ^  =  s>  +  2  xy, 

90.    Dependent  Variables.  —  It  often  happens  that  some 
of  the  variables  are  functions  of  others.     For  example,  let 

u  =  x2  +  y2  +  z2 

and  let  z  be  a  function  of  x  and  y.     When  y  is  constant,  z  will 

be  a  function  of  x  and  the  partial  derivative  of  u  with  respect 

to  x  will  be 

du  dz 

— -  =  2z  +  2  s— 
dx  dx 

Similarly,  the  partial  derivative  with  respect  to  y  with  x  con- 
stant is 

—  =  2y  +  2  z  —  > 
dy  "  dy 

If,  however,  we  consider  z  constant,  the  partial  derivatives 

are 

du  du 

—  =  2  x,     —  =  2  y. 

dx  '     dy  y 

The  value  of  a  partial  derivative  thus  depends  on  what  quantities 
are  kept  constant  during  the  differentiation. 

The  quantities  kept  constant  are  sometimes  indicated  by 
subscripts.     Thus,  in  the  above  example 

\dx)ytZ  '  \dxly  dx      \dxjz  J  dx 

*  For  a  proof  see  Wilson,  Advanced  Calculus,^  50. 


116 


DIFFERENTIAL  CALCULUS 


Chap.  XL 


It  will  usually  be  clear  from  the  context  what  independent 

Su 
variables  u  is  considered  a  function  of.     Then  —  will  repre- 

OX 

sent  the  derivative  with  all  those  variables  except  x  constant. 
Example.     If  a  is  a  side  and  A  the  opposite  angle  of  a  right 

B  triangle  with  hypotenuse  c,  find  (  — )    • 

\oc/a 

From  the  triangle  it  is  seen  that 
c  a  =  c  sin  A. 

Differentiating  with  A  constant,  we  get 

da 


Fig.  90. 


dc 


=  sm  A, 


which  is  the  value  required. 

91.  Geometrical  Representation.  —  Let  z  =  /  (x,  y)  be 
the  equation  of  a  surface.  The  points  with  constant  y- 
coordinate  form  the  curve  AB  (Fig.  91a)  in  which  the  plane 
y  =  constant  intersects  the  surface.  In  this  plane  z  is  the 
vertical  and  x  the  horizontal  coordinate.     Consequently, 

dz 

dx 

is  the  slope  of  the  curve  AB  at  P. 

Similarly,  the  locus  of  points  with  given  x  is  the  curve  CD 
and 

dz_ 

dy 

is  the  slope  of  this  curve  at  P. 

Example.     Find  the  lowest  point  on  the  paraboloid 

z  =  x2  +  y2_2x-4:y  +  6. 

At  the  lowest  point,  the  curves  AB  and  CD  (Fig.  91b)  will 
have  horizontal  tangents.     Hence 


^  =  2z-2  =  0,         |* 
dx  dy 


=  2  y  -  4  =  0. 


Chap.  XI. 


PARTIAL   DIFFERENTIATION 


117 


Consequently,  x  =  1,  y  =  2.  These  values  substituted  in 
the  equation  of  the  surface  give  z  =  1.  The  point  required 
is  then  (1,  2,  1).  That  this  is  really  the  lowest  point  is  shown 
by  the  graph. 


Fig.  91a. 


Fig.  91b. 


EXERCISES 

In  each  of  the  following  exercises  show  that  the  partial  derivatives 
satisfy  the  equation  given : 


L  t*  = 


a?  +  tf 


x  +  y' 

2.  z  =  (x  +  a)  (y  +  b), 

3.  z  =  Or2  +  y2)n, 


du  du  P 

dz  dz 


dx  dy 
dz 


=  2> 

dz 


«-  =  x  —  • ' 
J  dx  dy 


4.   u  =  In  Cr2  +  xy  +  y2),     xp  +  y^ 


dX 


dy 


6.    w  =  -  H h  -, 

2      x       ?/ 


6.    u  =  tan  J 


7.    w  = 


(-)• 


1 


<9m  ,       du    .       du       n  „ 
x  —  +  y  —  +  z  —  =  0. 
do;  d#  dz 

d-u   .    d2lf        „ 

d.r2  d//- 

a2M  a*ti  .  a*!*^  Q  ,/ 

dx2  d//2        dz2 


Vx2  +  ?/2  +  z2' 
In  each  of  the  following  exercises  verify  that 

d2u  d2u 

dx  dy       dy  dx 


118 


DIFFERENTIAL  CALCULUS 


Chap.  XI. 


8.  u  =  -.  10.    u  =  sin  (re  +  y)S 

x 

9.  u  =  In  (x2  +  2/2).  11.   it  =  xyz. 

12.  Given  y  =  Vx2  + 1/2  +  s2,  verify  that 

dzv  dzv 

dx  dy  dz      dz  By  dx 

Prove  the  following  relations  assuming  that  z  is  a  function  of  x  and  yi 

13.  u  =  (x  +  z)ev+*,     ~  +  ^  =  (l  +  z+*)(l+^  +  PW*. 


dx       dy 


dx      dyt 


I    du  du\         (    dz  dz\ 

li.u  =  xyz,    e\x--y-^u\x--y-y 

\dxdy      dx  dyj  J 


15.   u 
16. 


-  ey  +  e2, 
dz 


dx  dy  \  dx  dy 

d  I    du 

—  \  z  — ■  — 

dx  \     dX  dX 

17.    If  x  =  r  cos  9,  y  =  r  sin  0,  show  that 


\         d2u 
)  ~  Z  dx2 


d2z  , 
dx2 


fdx\        (drY 
\dr)9=:\dx)y 


18.  Let  a  and  b  be  the  sides  of  a  right  triangle  with  hypotenuse  c  and 
opposite  angles  A  and  B,  Let  p  be  the  perpendicular  from  the  vertex 
of  the  right  angle  to  the  hypotenuse.     Show  that 

fdp\   =  ¥  (djp\     _  b 

\dajb~  c3'  \dajA~c' 

19.  If  K  is  the  area  of  a  triangle,  a  side  and  two  adjacent  angles  of 
which  are  c,  A,  B,  show  that 

dK\       =  &         fdK\       =o* 


c 


f 


<ta/c,5         2'  \dBjc,A        2 

20.  If  X  is  the  area  of  a  triangle  with  sides  a,  b,  c,  show  that 

dK\  a       .    . 

—  =  -  cot  A. 

da  )b,c      2 

21.  Find  the  lowest  point  on  the  surface 

z  =  2  x2  +  y2  +  8  x  -  2  ?/  +  9. 

22.  Find  the  highest  point  on  the  surface 

z  =  2  y  -  x2  +  2  xy  -  2  y*  +  I. 

92.  Increment.  —  Let  u  =  f  (x,  y)  be  a  function  of  two 
independent  variables  x  and  y.  When  x  changes  to  x  +  Ax 
and  y  to  y  -\-  Ay,  the  increment  of  u  is 

Aw  =  /  (x  +  Ax,  2/  +  A?/)  -  /  (X,  y).  (92a) 


Chap.  XI.  PARTIAL   DIFFERENTIATION  119 

By  the  mean  value  theorem,  Art.  76, 

/  (x  +  Ax,  y  +  Ay)  =f(x,y  +  Ay)  +  Ax/X  (xb  y  +  Ay), 

X\  lying  between  £  and  x  +  Ax.     Similarly 

/  0,  y  +  Ay)  =  f  (x,  y)  +  Ayfy  (x,  yi), 

2/i  being  between  y  and  y  -f-  Ay.  Using  these  values  in  (92a), 
we  get 

Au  =  Axfx  (xh  y  +  Ay)  +  Ay/y  (x,  yi).  (92b) 

As  Ax  and  Ay  approach  zero,  Xi  approaches  x  and  yi  ap- 
proaches y.     If  /x  (x,  y)  and  /y  (x,  y)  are  continuous, 

du 
fx  fa,  y  +  Ay)  =  U  fo 2/)  +  «i  =  ^  +  €i, 

du 
fy  (x,  Vl)  =  fy  (x,  y)  +  €2  =  g-  +  €2, 

€i  and  62  approaching  zero  as  Ax  and  Ay  approach  zero. 
These  values  substituted  in  (92b)  give 

Au  =  ^  Ax  +  ~  Ay  +  €i  Ax  +  e2  Ay.  (92c) 

The  quantity 

du  A  ch£  . 

—  Ax  +  —  Ay 
dx  dy 

is  called  the.  principal  part  of  Aw.  It  differs  from  Au  by  an 
amount  ei  Ax  +  e2  Ay.  As  Ax  and  Ay  approach  zero,  €i  and  e2 
approach  zero  and  so  this  difference  becomes  an  indefinitely 
small  fraction  of  the  larger  of  the  increments  Ax  and  Ay. 
We  express  this  by  saying  the  principal  part  differs  from  Au 
by  an  infinitesimal  of  higher  order  than  Ax  and  Ay  (Art.  9). 
When  Ax  and  Ay  are  sufficiently  small  this  principal  part  then 
gives  a  satisfactory  approximation  for  Au. 

Analogous  results  can  be  obtained  for  any  number  of  in- 
dependent variables.  For  example,  if  there  are  three  inde- 
pendent variables  x,  y,  z,  the  principal  part  of  Au  is 

du  .      .  du  A      ,  du  A 
-AZ+-AJ/  +  -A*. 

In  each  case,  if  the  partial  derivatives  are  continuous,  the 


120  DIFFERENTIAL  CALCULUS  Chap.  XI. 

principal  part  differs  from  Au  by  an  amount  which  becomes 
indefinitely  small  in  comparison  with  the  largest  of  the  in- 
crements of  the  independent  variables  as  those  increments 
all  approach  zero. 

Example.  Find  the  change  in  the  volume  of  a  cylinder 
when  its  length  increases  from  6  ft.  to  6  ft.  1  in.  and  its  diam- 
eter decreases  from  2  ft.  to  23  in. 

Since  the  volume  is  v  =  irr2h,  the  exact  change  is 

AV  =  7T  (1   -   v\)2  (6  +  TV)    -  7T  •  I2  •  6   =    -0.413  7T  cu.   ft. 

The  principal  part  of  this  increment  is 

ZAr+diAh  =  2*rh(-k)+rr2  (i) = -°-417  * cu  ft- 

93.  Total  Differential.  —  If  u  is  a  function  of  two  inde- 
pendent variables  x  and  y,  the  total  differential  of  u  is  the 
principal  part  of  Aw,  that  is, 

du  du 

du  =  —  Ax  +  —  Ay.  (93a) 

dx  dy 

This  definition  applies  to  any  function  of  x  and  y.  The 
particular  values  u  =  x  and  u  =  y  give 

dx  =  Ax,        dy  =  Ay,  (93b) 

that  is,  the  differentials  of  the  independent  variables  are  equal 
to  their  increments. 

Combining  (93a)  and  (93b),  we  get 

du  =  —  dx  +  —  dy.  (93c) 

dx  dy 

We  shall  show  later  (Art.  97)  that  this  equation  is  valid  even 

if  x  and  y  are  not  the  independent  variables. 

The  quantities 

7  du   7  ,  du  , 

dxu  =  —  dx,        dyu  =  —  dy 
dx  dy    * 

are  called  partial  differentials.  Equation  (93c)  expresses 
that  the  total  differential  of  a  function  is  equal  to  the  sum  of  the 
partial  differentials  obtained  by  letting  the  variables  change  one 
at  a  time. 


Chap.  XL  PARTIAL   DIFFERENTIATION  121 

* 

Similar  results  can  be  obtained  for  functions  of  any  number 
of  variables.  For  instance,  if  u  is  a  function  of  three  inde- 
pendent variables  x,  y,  z, 

1  du    .  du   A        .    du    . 

du  -  -r-  Ax  +  t-  Ay  +  -T-  Az. 

d£  dy  dz 

The  particular  values  u  =  x,  u  =  y,  u  =  z  give 
dx  =  A.r,         dy  =  Ay,        dz  =  Az. 
The  previous  equation  can  then  be  written 

du  =  P  dx  +  ^  dy  +  ^  dz  (93d) 

dx  dy  dz 

and  in  this  form  it  can  be  proved  valid  even  when  x,  y,  z  are 
not  the  independent  variables. 

Example  1.     Find  the  total  differential  of  the  function 

u  =  x2y  +  xy2. 

By  equation  (93  c) 

,         du  ,         du  7 
aw  =  —  az  +  —  ay  . 
dx  dy 

=  (2xy  -\-  y2)  dx  +  (x2  +  2  xy)  dy. 

Ex.  2.  Find  the  error  in  the  volume  of  a  rectangular  box 
due  to  small  errors  in  its  three  edges. 

Let  the  edges  be  x,  y,  z.     The  volume  is  then 

v  =  xyz. 

The  error  in  v,  due  to  small  errors  \x,  Ay,  Az  in  x,  y,  z,  is  Av. 
If  the  increments  are  sufficiently  small,  this  will  be  approxi- 
mately 

dv  =  yzdx  +  xz  dy  +  xy  de. 

Dividing  by  v,  we  get 

dv      yz  dx  -f-  xzdy  -f  #y  a*g 

as      J//      dz 

= 1 

xyz 

dx 
Now  —  expresses  the  error  dx  as  a  fraction  or  percentage  of  x. 


122  DIFFERENTIAL  CALCULUS  Chap.  XI. 

The  equation  just  obtained  expresses  that  the  percentage 
error  in  the  volume  is  equal  to  the  sum  of  the  percentage 
errors  in  the  edges.  If,  for  example,  the  error  in  each  edge 
is  not  more  than  one  per  cent,  the  error  in  the  volume  is  not 
more  than  three  per  cent. 

94.  Calculation  of  Differentials.  —  In  proving  the  formu- 
las of  differentiation  it  was  assumed  that  u,  v}  etc.,  were 
functions  of  a  single  variable.  It  is  easy  to  show  that  the 
same  formulas  are  valid  when  those  quantities  are  functions 
of  two  or  more  variables  and  du,  dv,  etc.,  are  their  total 
differentials. 

Take,  for  example,  the  differential  of  uv.  By  (93c)  the 
result  is 

d  (uv)  =  —  (uv)  du  +  —  (uv)  dv  =  v  du  +  u  dv, 

du  dv 

which  is  the  formula  IV  of  Art.  17. 
Example,     u  =  yex  +  zey. 
Differentiating  term  by  term,  we  get 

du  =  yex  dx  +  ex  dy  +  zev  dy  +  ev  dz. 

We  obtain  the  same  result  by  using  (93d) ;  for  that  formula 
gives 

du  du  du 

du  —  —  dx  +  t—  dy  +  —  dz  =  ye  dx  +  (ex  +  zev)  dy  -f-  ev  dz. 
dx  dy  dz 

95.  Partial  Derivatives   as   Ratios   of   Differentials.  — 

The  equation 

,  du  1 

dxu  =  —  dx 
dx 

du 
shows  that  the  partial  derivative  —  is  the  ratio  of  two  dif- 

ox 

ferentials  dxu  and  dx.     Now  dxu  is  the  value  of  du  when  the 

same  quantities  are  kept  constant  that  are  constant  in  the 

calculation  of  -r-.     Therefore,  the  partial  derivative  —  is  the 
dx  dx 


Chap.  XI.  PARTIAL   DIFFERENTIATION  123 

du 
value  to  which  -s-  reduces  when  du  and  dx  are  determined  with 
dx 

the  same  quantities  constant  that  are  constant  in  the  calculation 

.  du 

of—- 

dx 

Example.     Given  u  =  x2  +  y2  +  z2}  v  =  xyz,  find  f  —  J     • 

Differentiating  the  two  equations  with  v  and  z  constant, 

we  get 

du  =  2x  dx  -\-  2y  dy,        0  =  yz  dx  +  xz  dy. 

Eliminating  dy, 


du  =  2  x  dx  -  2  2-  dx  =  2  ' 


J  dx. 


x  \     x 

Under  the  given  conditions  the  ratio  of  du  to  dx  is  then 

du      2  (x2  —  y2) 
dx  x 

Since  v  and  z  were  kept  constant,  this  ratio  represents  ( —  J    ; 

that  is, 

fdu\  2  (x2  —  y2) 


\OX  J  v tz  X 

EXERCISES 

1.  One  side  of  a  right  triangle  increases  from  5  to  5.2  while  the  other 
decreases  from  12  to  11.75.  Find  the  increment  of  the  hypotenuse  and 
its  principal  part. 

2.  A  closed  box,  12  in.  long,  8  in.  wide,  and  6  in.  deep,  is  made  of 
material  \  inch  thick.  Find  approximately  the  volume  of  material 
used. 

^.   Two  sides  and  the  included  angle  of  a  triangle  are  b  =  20,  c  =  30, 
and  A  =  45°.     By  using  the  formula  f 

a2  =  b-  +  c2  -  2  be  cos  A,       ♦* 

find  approximately  the  change  in  a  when  6  increases  1  unit,  c  decreases 
\  unit,  and  A  increases  1  degree. 
4.    The  period  of  a  simple  pendulum  is 

T  =  2  tt  \/!  • 
v  9 

Find  the  error  in  T  due  to  small  errors  in  I  and  g. 


124  DIFFERENTIAL   CALCULUS  Chap.  XI. 

5.  If  g  is  computed  by  the  formula, 

s  =  2  gt2, 
find  the  error  in  g  due  to  small  errors  in  s  and  t. 

6.  The  area  of  a  triangle  is  determined  by  the  formula 

K  =  \  ab  sin  C. 

Find  the  error  in  K  due  to  small  errors  in  a,  b,  C. 

Find  the  total  differentials  of  the  following  functions: 

7.  xy2z\  9.    ?  +  £  +  •?. 

y      z      x 

8.  xy  sin  (x  +  y).  10.    tan-1  -  -f-  tan-1  — 

*  *'  x  y 

11.  The  pressure,  volume,  and  temperature  of  a  perfect  gas  are  con- 
nected by  the  equation  pv  =  kt,  k  being  constant.  Find  dp  in  terms  of 
dv  and  dt. 

12.  If  x,  y  are  rectangular  and  r,  0  polar  coordinates  of  the  same 
point,  show  that 

x  dy  —  y  dx  =  r2  dd,     dx2  +  dy2  =  dr2  +  r2  dd2.  ' 


13.  If  a:  =  u  —  v,  y  =  u2  +  v2,  find  f  ~  J 

14.  If  u  =  xy  +  yz  +  z#,  a;2  +  z2  =  2  ?/z,  find  [  —  j   • 

\dZ  jy 


'¥)■ 


15.    If  yz  =  ux  +  ?>2,  vx  =  uy  +  z2,  find 


fdv\ 

\dzju,x 


16.    A  variable  triangle  with  sides  a,  6,  c  and  opposite  angles  A,  B,  C 
is  inscribed  in  a  fixed  circle.     Show  that 

da  db  dc 

+  — — ~b  +  ZZTF,  =  v-K 


cos  A       cos  B      cos  C 

96.   Derivative  of  a  Function  of  Several  Variables.  — 

Let  u  =  f  (x,  y)  and  let  x  and  y  be  functions  of  two  variables 
s  and  t.  When  t  changes  to  t  +  At,  x  and  y  will  change  to 
x  +  Ax  and  y  +  Ay.     The  resulting  increment  in  u  will  be 

Aw  =  -—  Ax  +  —  A?/  +  ci  Ax  +  €2  Ay. 
.    ox  ay 

Consequently, 

Au      du  Ax      du  Ay         Ax         Ay 
At  ~  dx  At  +  dy  At  +  *  A«  +  *  A* 

As  AZ  approaches  zero,  Ax  and  Ay  will  approach  zero  and  so 


Chap.  XI.  PARTIAL   DIFFERENTIATION  125 

€i  and  €2  will  approach  zero.     Taking  the  limit  of  both  sides, 

du      du  dx       du  dy  ,       . 

dt  ~  di  ~dt  +  dy  Tt '  (     l) 

dx 
If  x  or  y  is  a  function  of  t  only,  the  partial  derivative  — 

dt 

d  if  dx       du 

or  -rj  is  replaced  by  a  total  derivative  -r±  or  -y- .     If  both  x 
dt  dt        dt 

and  y  are  functions  of  t,  u  is  a  function  of  £  with  total  deriva- 
tive 

du      du  dx      du  dy  (Qau\ 

dt~dx'di  +  dy"di'  {™} 

Likewise,  if  u  is  a  function  of  three  variables  x,  y,  z,  that 
depend  on  t, 

du       du  dx       du  dy       du  dz  ,Q     . 

'dt~dx"di  +  dy'dt  +  dz~di'  ^       ' 

As  before,  if  a  variable  is  a  function  of  t  only,  its  partial  de- 
rivative is  replaced  by  a  total  one.  Similar  results  hold  for 
any  number  of  variables. 

The  term 

du  dx 

dx  dt 

is  the  result  of  differentiating  u  with  respect  to  t,  leaving  all 
the  variables  in  u  except  x  constant.  Equations  (96a)  and 
(96c)  express  that  if  u  is  a  function  of  several  variable  quanti- 

tics,  —  can  be  obtained  by  differentiating  with  respect  to  t  as  if 

only  one  of  those  quantities  were  variable  at  a  time  and  adding 
the  results. 

7 

Example  1.    Given  y  =  xx,  find  -~- 

The  function  x*  can  be  considered  a  function  of  two  vari- 
ables, the  lower  x  and  the  upper  x.  If  the  upper  x  is  held 
constant  and  the  lower  allowed  to  vary,  the  derivative  (as  in 
case  of  xn)  is 

x  .  xx~l  =  x*. 


126  DIFFERENTIAL  CALCULUS  Chap.  XI. 

If  the  lower  x  is  held  constant  while  the  upper  varies,  the 
derivative  (as  in  case  of  ax)  is 

x*  \nx. 

The  actual  derivative  of  y  is  then  the  sum 

■j-  =  xx  +  xx  In  x. 

Ex.  2.   Given  u  =  f  (x,y,  z),  y  and  z  being  functions  of  x, 

n    ,  du 

find  7Z' 
ax 

By  equation  (96c)  the  result  is 

du      du      du  dy      du  dz 
dx      dx      dy  dx       dz  dx 

In  this  equation  there  are  two  derivatives  of  u  with  respect 

to  x.     If  y  and  z  are  replaced  by  their  values  in  terms  of  x,  u 

will  be  a  function  of  x  only.     The  derivative  of  that  function 

du 
is  -T- .     If  y  and  z  are  replaced  by  constants,  u  will  be  a  second 

du 
function  of  x.     Its  derivative  is  —  • 

dx 

Ex.  3.  Given  u  =  f  (x,  y,  z),  z  being  a  function  of  x  and  y. 
Find  the  partial  derivative  of  u  with  respect  to  x. 

It  is  understood  that  y  is  to  be  constant  in  this  partial 
differentiation.     Equation  (96c)  then  gives 

du      du      du  dz 
dx       dx       dz  dx 

In  this  equation  appear  two  partial  derivatives  of  u  with 
respect  to  x.  If  z  is  replaced  by  its  value  in  terms  of  x  and  y, 
u  will  be  expressed  as  a  function  of  x  and  y  only.  Its  partial 
derivative  is  the  one  on  the  left  side  of  the  equation.  If  z  is 
kept  constant,  u  is  again  a  function  of  x  and  y.  Its  partial 
derivative  appears  on  the  right  side  of  the  equation.  We 
must  not  of  course  use  the  same  symbol  for  both  of  these 
derivatives.     A  way  to  avoid  the  confusion  is  to  use  the 


Chap.  XI.  PARTIAL   DIFFERENTIATION  127 

letter/  instead  of  u  on  the  right  side  of  the  equation.  It  then 
becomes 

du  =  d£      d£  dz^ 

dx       dx       dz  dx 

It  is  understood  that/  (x,  y}  z)  is  a  definite  function  of  x,  y,  z 

df 

and  that  -^-  is  the  derivative  obtained  with  all  the  variables 
dx 

but  x  constant. 

97.  Change  of  Variable.  —  If  u  is  a  function  of  x  and  y 
we  have  said  that  the  equation 

7         du   7         du   7 
du  =  —  dx  +  —  dy 
dx  dy 

is  true  whether  x  and  y  are  the  independent  variables  or  not. 
To  show  this  let  s  and  t  be  the  independent  variables  and  x 
and  y  functions  of  them.     Then,  by  definition, 

,  dll    -.      .    dll    ,, 

du  =  —  ds  +  —  dt. 
ds  dt 

Since  u  is  a  function  of  x  and  y  which  are  functions  of  s  and  t, 
by  equation  (96a), 

dll       dll  dx       du  dy  du       du  dx       dll  dy^ 

ds~dxdsdydsJ  dt  ~  dx  Hi       dy  dt' 

Consequently, 

/du  dx       dll  dy\  ,      .    /dll  dx    .    du  dlj\  ,, 

du=  \TxTs  +  TyTsr  +  \TxTt+TyTtr 

=  *JL(fds  +%dt)  +  £&*+%  dt)  =  ^dx  +  pdy, 
dx\ds  dt      /       dy\ds  dt      J      dx  dy 

which  was  to  be  proved. 

A  similar  proof  can  be  given  in  case  of  three  or  more 
variables. 

98.  Implicit  Functions.  —  If  two  or  more  variables  are 
connected  by  an  equation,  a  differential  relation  can  be  ob- 
tained by  equating  the  total  differentials  of  the  two  sides  of 
the  equation. 


128  DIFFERENTIAL  CALCULUS  Chap.  XL 

Example  1.  /  (x,  y)  —  0. 
In  this  case 

d-f(x,y)  =^dx  +  ^dy^d. 0  =  0.\ 
Consequently, 

dy  dx 

dx  df 

dy 

Ex.  2.  /  (x,  y,  z)  =  0. 
Differentiation  gives 

4-  dx  +  -f-  dy  +  ~  dz  =  0, 

dx  dy  dz 

If  z  is  considered  a  function  of  x  and  y,  its  partial  derivative 
with  respect  to  x  is  found  by  keeping  y  constant.  Tl<en 
dy  =  0  and 

cte  dx 

dx  ~  ~~df 
dz 

Similarly,  if  x  is  constant,  dx  =  0  and 

dz  dy 

dz 

Ex.  3.  /i  (re,  I/,  z)  =  0,  /2  (z,  ?/,  2!)  =  0.    # 
We  have  two  differential  relations 

dx  dy  dz 

^dx  +  pdy+^dz-O. 

dx  dy  dz 

We  could  eliminate  y  from  the  two  equations  /i  =  0,  ji  =  0. 
We  should  then  obtain  z  as  a  function  of  x.     The  total  de- 


Chap.  XI. 


PARTIAL  DIFFERENTIATION 


129 


rivative  of  this  function  is  found  by  eliminating  dy  and  solving 

dz 
for  the  ratio  -j-  •     The  result  is 
ax 


dz 


dy  dx       dx  dy 


dx"  dj1dj1_df1dh 
dz  dy       dy   dz 

99.  Directional  Derivative.  —  Let  u  =  f  (x,  y).  At  each 
point  P  (x,  y)  in  the  z?/-plane,  u  has  a  definite  value.  If  we 
move  away  from  P  in  any  definite  direction  PQ,  x  and  y  will 


o 


Fig.  99. 

be  functions  of  the  distance  moved.     The  derivative  of  u 
with  respect  to  s  is 

du      du  dx   .   da  dy       du  du    . 

—  = ^  +  t 7^  =  ^"  cos  <b  +  t-  sin  </>. 

ds       dx  ds       dy  ds       dx  dy 

This  is  called  the  derivative  of  u  in  the  direction  PQ.     The 

partial  derivatives  —  and  —  are  special  values  of  —  which 
1  dx  dy  ds 

result  when  PQ  is  drawn  in  the  direction  of  OX  or  OY. 

Similarly,  if  u  =  f  (x,  y,  z), 

du      dudx  .   dudy  .   du  dz      du  du  du 

—  =  —  t-  +  t-  -r  +  v-  -r  =  T~  cos  o;  +  -T-  cos  /3  +  -r-  cos  7 

ds       dx  ds       dy  ds       dz  ds       dx  dy  dz 

is  the  rate  of  change  of  u  with  respect  to  s  as  we  move  along  a 
line  with  direction  cosines  cos  a,  cos  0,  cos  y.     The  partial 


130  DIFFERENTIAL  CALCULUS  Chap.  XI. 

< 

du 
derivatives  of  u  are  the  values  to  which  —  reduces  when  s  is 

ds 

measured  in  the  direction  of  a  coordinate  axis. 

Example.     Find  the  derivative  of  x2  +  y2  in  the  direction 
tf>  =  45°  at  the  point  (1,  2). 

The  result  is 

~(x2  +  y2)  =  2x^+2y  p-  =  2xcos<f>  +  2ysin<l> 

OS  OS  OS 

V2  V2 

100.   Exact  Differentials.  —  If  P  and  Q  are  functions  of 
two  independent  variables  x  and  y, 

Pdx  +  Qdy 

may  or  may  not  be  the  total  differential  of  a  function  u  oix 
and  y.     If  it  is  the  total  differential  of  such  a  function,      ^ 

P  dx  -\-  Q  dy  =  du  =  —  dx  -\-  —  dy. 
^    *  dz  dy    * 

Since  d#  and  d?/  are  arbitrary,  this  requires 


P  = 

du 
dx' 

Q- 

du 
dy 

Consequently, 

dP 

d2u 

dQ_ 

d2u 

dy 

dy  dx' 

dx 

dxdy 

Since  the  two  second  derivatives  of  u  with  respect  to  x  and 
y  are  equal, 

%  -  S- 

An  expression  P  dx  -\-  Qdy  \s  called  an  exact  differential  \^/ 
it  is  the  total  differential  of  a  function  of  x  and  y.     We  have 
just  shown  that  (100a)  must  then  be  satisfied.     Conversely, 
it  can  be  shown  that  if  this  equation  is  satisfied  P  dx  +  Q  dy 
is  an  exact  differential.* 

*  See  Wilson,  Advanced  Calculus,  §  92. 


Chap.  XI.  PARTIAL  DIFFERENTIATION  131 

Similarly,  if 

P  dx  +  Q  dy  +  Rdz 

is  the  differentia]  of  a  function  u  of  x,  y,  z, 

dP  =  dQf         dQ  =  dRf         §E^?Et         /100b) 

dy        dx'  dz        dy'  dx        dz  ' 

and  conversely. 

Example  1.    Show  that 

(.r2  +  2  xy)  dx  +  (x2  +  y2)  dy 

is  an  exact  differential. 
In  this  case 

The1  two  partial  derivatives  being  equal,  the  expression  is 
exact. 

Ex,  2.    In  thermodynamics  it  is  shown  that 

dU  =  TdS  -  pdv, 

U  being  the  internal  energy,  T  the  absolute  temperature,  S 
the  entropy,  p  the  pressure,  and  v  the  volume  of  a  homogene- 
ous substance.  Any  two  of  these  five  quantities  can  be 
assigned  independently  and  the  others  are  then  determined. 
Show  that 

\dpJs      \dSjp 

The  result  to  be  proved  expresses  that 

TdS  +  vdp 

is  an  exact  differential.  That  such  is  the  case  is  shown  by 
replacing  T  dS  by  its  value  dU  +  p  dv.    We  thus  get 

TdS  +  vdp  =  dU  +  p  dv  +  v  dp  =  d  (U  +  pv). 

EXERCISES 
1.    If  u  =  f  (X,  y),  y  =  <t>  (x),  find  ^« 


2.    If  u  =  f  (x,  y,  z),  z  =  4>  (x),  find  f  —  j 


132  DIFFERENTIAL   CALCULUS  Chap.  XI. 

du 

3.  If  u  =  /  (x,  y,  z),  z  =  <i>  (x,  y),  y  =  $  (x),  find  ~^ 

4.  Uu=f(x,y),    y  =  4>(x,r),    r  =  rp(x,s),    find   U|J  ,  (—)  , 

«•  (a- 

5.  If  /  (x,  y,  2)  =  0,  Z  =  F  (x,  y),  findg- 

6.  If  F  (x,  y,  z)  =  0,  show  that 

dx  dy  dz 
dy  dz  dx 

7.  If  u  =  xf  (z),  z  =  -,  show  that  x- — \-  y  —  =  u. 

Sit  Oil  du 

8.  If  u  =f  (r,  s),  r  =  x-\-at,  s  =y  +  bt,  show  that  —  =  a  - — f-  b  — • 

dt  dx  dy 

9.  If  z  =  f  (x  +  ay),  show  that  —  =  a  — — 

dy  dx 

10.  li  u  =  f  {x,  y),  x  =  r  cos  Q,  y  =  r  sin  0,  show  that 

\  ^r  /  +  \r    30/         W/        U/// 

11.  The  position  of  a  pair  of  rectangular  axes  moving  in  a  plane  is 
determined  by  the  coordinates  h,  k  of  the  moving  origin  and  the  angle  <t> 
between  the  moving  .r-axis  and  a  fixed  one.  A  variable  point  P  has  co- 
ordinates x',  y'  with  respect  to  the  moving  axes  and  x,  y  with  respect  to 
the  fixed  ones.     Then 

x  =  /  0',  y',  h,  k,4>),        y  =  F  (x',  y',  h,  k,  <f>). 

Find  the  velocity  of  P.  Show  that  it  is  the  sum  of  two  parts,  one  repre- 
senting the  velocity  the  point  would  have  if  it  were  rigidly  connected 
with  the  moving  axes,  the  other  representing  its  velocity  with  respect 
to  those  axes  conceived  as  fixed. 

12.  Find  the  directional  derivatives  of  the  rectangular  coordinates 
x,  y  and  the  polar  coordinates  r,  d  of  a  point  in  a  plane.  Show  that  they 
are  identical  with  the  derivatives  with  respect  to  s  given  in  Arts.  54  and 
59. 

13.  Find  the  derivative  of  x2  —  y2  in  the  direction  4>  =  30°  at  the 
point  (3,  4). 

14.  At  a  distance  r  in  space  the  potential  due  to  an  electric  charge  e 

is  V  =  - .     Find  its  directional  derivative. 
r 

15.  Show  that  the  derivative  of  xy  along  the  normal  at  any  point  of 
the  curve  x2  —  y2  =  a2  is  zero. 

V 


4# 


Chap.  X  I . 


PARTIAL  DIFFERENTIATION 


133 


16.    Given  u  =  /  (x,  y),  show  that 


\dsy)  +  \0s,)      :  \dx)  +  \0y 


)■■ 


if  si  and  9|  are  measured  along  perpendicular  directions. 

Determine  which  of  the  following  expressions  are  exact  differentials: 

17.  \j  dx  —  x  dy. 

18.  (2  x  +  y)  dx  +  (x  -  2  y)  dy. 

19.  eg  dx  +  cy  dy  +  {x  -\-  y)  ez  dz. 

20.  yz  dx  —  xz  dy  +  \f  dz. 

21.  Under  the  conditions  of  Ex.  2,  page  131,  show  that 

(dv\  (dS\  /0p\         (d§\ 

\dTjp-        {dpJT         \dT)v~    \d»  )t 

22.  In  case  of  a  perfect  gas,  pv  =  kT.     Using  this  and  the  equation 

dU  =  TdS  -p  dv, 


show  that 


^  =  0. 

dp 


Since  U  is  always  a  function  of  p  and  T,  this  last  equation  expresses 
that  U  is  a  function  of  T  only. 

101.   Direction  of  the  Normal  at  a  Point  of  a  Surface.  — 
Let  the  equation  of  a  surface  be 

F  (x,  y,  z)  =  0. 
Differentiation  gives 


dx  dy 

Let  PN  be  the  line 
through  P  (x,  y,  z)  with 
direction  cosines  propor- 
tional to 

dF^dFdF 
dx  '  dy      dz 

If  P  moves  along  a  curve 
on  the  surface,  the  direc- 
tion cosines  of  its  tangent 
PT  are  proportional  to 
dx  :  dy  :  dz. 


—  dx  ■    — dy  +  -7-  dz      0. 


(101a) 


Fig.  101. 


Equation  (101a)  expresses  that  PN  and  PT  are  perpendicu- 
lar to  each  other  (Art.  61).     Consequently  PN  is  perpendicu- 


134 


DIFFERENTIAL  CALCULUS 


Chap.  XI. 


lar  to  all  the  tangent  lines  through  P.  This  is  expressed  by 
saying  PN  is  the  normal  to  the  surface  at  P.  We  conclude 
that  the  normal  to  the  surface  F  (x,  y,  z)  =  0  at  P  (x,  y,  z)  has 
direction  cosines  proportional  to 

—  •—•—.  (101) 

dx  '  dy  '  dz 

102.  Equations  of  the  Normal  at  Pi  {oc\,  y\,  z{).  —  Let  A, 

B,  C  be  proportional  to  the  direction  cosines  of  the  normal 

to  a  surface  at  Pi  (xi,  yh  zi).     The  equations  of  the  normal 

are  (Art.  63) 

x-xi_y-yi      z  -  Zj 

—AT=~B~  =  ^~  (102) 

103.  Equation  of  the  Tangent  Plane  at  Pi  (a>i,  yl9  zi).  — 
All  the  tangent  lines  at  Pi  on  the  surface  are  perpendicular 


Fig.  103. 

to  the  normal  at  that  point.  All  these  lines  therefore  lie  in  a 
plane  perpendicular  to  the  normal,  called  the  tangent  plane 
at  Pi. 

It  is  shown  in  analytical  geometry  that  if  A ,  B,  C  are  pro- 
portional to  the  direction  cosines  of  the  normal  to  a  plane 
passing  through  (xi,  yu  zi),  the  equation  of  the  plane  is 

A(x-xi)  +  B  (y  -yi)  +  C(z-  z4)  =  0.*      (103) 

*  See  Phillips,  Analytic  Geometry,  Art.  68. 


Chap.  XI.  PARTIAL   DIFFERENTIATION  135 

If  A,  B,  C  are  proportional  to  the  direction  cosines  of  the 
normal  to  a  surface  at  Pi,  this  is  then  the  equation  of  the 
tangent  plane  at  Pi. 

Example.  Find  the  equations  of  the  normal  line  and  tan- 
gent plane  at  the  point  (1,  —1,  2)  of  the  ellipsoid 

x2  +  2  y2  +  3  z2  =  3  x  +  12. 

The  equation  given  is  equivalent  to 

x2  +  2  if  +  3  z2  -  3  x  -  12  =  0. 

The  direction  cosines  of  its  normal  are  proportional  to  the 

partial  derivatives 

2z-3:4?/:6z. 

At  the  point  (1,  —1,  2),  these  are  proportional  to 
A  :B  :C  =  -1:  -4:  12  =  1:4:  -12. 

The  equations  of  the  normal  are 

x-l_y+l_z-2 
1      ~      4  -12  ' 

The  equation  of  the  tangent  plane  is 

x  -  1  +  4  (y  +  1)  -  12  (3  -  2)  =  0. 

EXERCISES 

Find  the  equations  of  the  normal  and  tangent  plane  to  each  of  the 
following  surfaces  at  the  point  indicated: 

1.  Sphere,  x2  +  y2  +  z2  =  9,  at  (1,  2,  2). 

2.  Cylinder,  x2  +  xy  +  if  =  7,  at  (2,  -3,  3). 

3.  Cone,  z2  =  x2  +  y2,  at  (3,  4,  5). 

4.  Hyperbolic  paraboloid,  xy  =  3  z  —  4,  at  (5,  1,  3). 

5.  Elliptic  paraboloid,  x  =  2  y2  +  3  z2,  at  (5,  1,  1). 

6.  Find  the  locus  of  points  on  the  cylinder 

Or  +  z)2  +  (,,  -  z)2  =  4 

where  the  normal  is  parallel  to  the  :r?/-plane. 

7.  Show  that  the  normal  at  any  point  P  (x,  y,  z)  of  the  surface 
y2  -f-  z2  =  4  x  makes  equal  angles  with  the  a>axis  and  the  line  joining 
P  and  A  (1,  0,  0). 

8.  Show  that  the  normal  to  the  spheroid 

x2  +  z2       y2 
9      "*"  25 

at  P  (x,  y,  z)  determines  equal  angles  with  the  lines  joining  P  with 
A'(0,  -4,0)  and  A  (0,4,0). 


136  DIFFERENTIAL   CALCULUS  Chap.  XI. 

104.  Maxima  and  Minima  of  Functions  of  Several 
Variables.  —  A  maximum  value  of  a  function  u  is  a  value 
greater  than  any  given  by  neighboring  values  of  the  variables. 
In  passing  from  a  maximum  to  a  neighboring  value,  the  func- 
tion decreases,  that  is 

ku  <  0.  (104a) 

A  minimum  value  is  a  value  less  than  any  given  by  neigh- 
boring values  of  the  variables.  In  passing  from  a  minimum 
to  a  neighboring  value 

Lu  >  0.  (104b) 

If  the  condition  (104a)  or  (104b)  is  satisfied  for  all  small 

changes  of  the  variables,  it  must  be  satisfied  when  a  single 

variable  changes.     If  then  all  the  independent  variables  but 

x  are  kept  constant,  u  must  be  a  maximum  or  minimum  in  x. 

3u 
If  —  is  continuous,  by  Art.  31, 
ox 

Therefore,  if  the  first  partial  derivatives  of  u  with  respect  to  the 
independent  variables  are  continuous,  those  derivatives  must  be 
zero  when  u  is  a  maximum  or  mini?num. 

When  the  partial  derivatives  are  zero,  the  total  differential 
is  zero.  For  example,  if  x  and  y  are  the  independent  vari- 
ables, 

du  =^dx+^fdy  =  0-dx  +  0-dy  =  0.      (104d) 

Therefore,  if  the  first  partial  derivatives  are  continuous,  the 
total  differential  of  u  is  zero  when  u  is  either  a  maximum  or  a 
minimum. 

To  find  the  maximum  and  minimum  values  of  a  function, 
we  equate  its  differential  or  the  partial  derivatives  with  re- 
spect to  the  independent  variables  to  zero  and  solve  the 
resulting  equations.  It  is  usually  possible  to  decide  from 
the  problem  whether  a  value  thus  found  is  a  maximum, 
minimum,  or  neither. 


Chap.  XI.  PARTIAL   DIFFERENTIATION  137 

Example  1.  Show  that  the  maximum  rectangular  parallele- 
piped with  a  given  area  of  surface  is  a  cube. 

Let  x,  y,  z  be  the  edges  of  the  parallelopiped.  If  V  is  the 
volume  and  A  the  area  of  its  surface 

V  =  xyz,        A  =  2  xy  +  2  xz  +  2  yz. 

Two  of  the  variables  x,  y,  z  are  independent.     Let  them  be 
x,  y.     Then 


Therefore 


A  —  2  xy 

"2(x  +  y) 


xy  (A  -  2  xy) 
2  (x  +  I/) 


67      x2[A  -  2y2  -  4  xy  1 
6y  "  2  L       («  +  y)2       J 
The  values  a;  =  0,  2/  =  0  cannot  give  maxima.     Hence 
i-2x2-4o;!/  =  0,         A  -  2y2  -  4xy  =  0. 
Solving  these  equations  simultaneously  with 

A  =  2xy  +  2xz  +  2yz, 
we  get 


# 


x  =  y  =  Z  =  y    g 

We  know  there  is  a  maximum.     Since  the  equations  give 
only  one  solution  it  must  be  the  maximum. 
Ex.  2.    Find  the  point  in  the  plane 

x  +  2  y  +  3  z  =  14 

nearest  to  the  origin. 

The  distance  from  any  point  (x,  y,  z)  of  the  plane  to  the 
origin  is 

D  =  Vx2  +  if  +  z2. 


138  DIFFERENTIAL  CALCULUS  Chap.  XI. 

If  this  is  a  minimum 

-,    r.      x  dx  +  y  dy  +  z  dz      ~ 
d  •  D  = u    u      =—  =  0, 

Vx2  +  2/2  +  ^2 
that  is, 

x  dx  +  y  dy  -{-  z  dz  =  0.  (104e) 

From  the  equation  of  the  plane  we  get 

dx  +  2dy  +  Zdz  =  0.  (104f) 

The  only  equation  connecting  z,  ?/,  2  is  that  of  the  plane. 
Consequently,  dx,  dy,  dz  can  have  any  values  satisfying  this 
last  equation.  If  x,  y,  z  are  so  chosen  that  D  is  a  minimum 
(104e)  must  be  satisfied  by  all  of  these  values.  If  two  linear 
equations  have  the  same  solutions,  one  is  a  multiple  of  the 
other.  Corresponding  coefficients  are  proportional.  The 
coefficients  of  dx,  dy,  dz  in  (104e)  are  x,  y,  z.  Those  in  (104f) 
are  1,  2,  3.     Hence 

1  ?  =  1  =  5 

12      3' 

Solving  these  simultaneously  with  the  equation  of  the  plane, 
we  get  x  =  1,  y  =  2,  z  =  3.  There  is  a  minimum.  Since 
we  get  only  one  solution,  it  is  the  minimum. 

EXERCISES 

1.  An  open  rectangular  box  is  to  have  a  given  capacity.  Find  the 
dimensions  of  the  box  requiring  the  least  material.    iT*  f%  %  V- 

2.  A  tent  having  the  form  of  a  cylinder  surmounted  by  a  cone  is  to 
contain  a  given  volume.  Find  its  dimensions  if  the  canvas  required  is  a 
minimum. 

3.  When  an  electric  current  of  strength  /  flows  through  a  wire  of 
resistance  R  the  heat  produced  is  proportional  to  PR.  Two  terminals 
are  connected  by  three  wires  of  resistances  Ri,  R2,  R3  respectively.  A 
given  current  flowing  between  the  terminals  will  divide  between  the 
wires  in  such  a  way  that  the  heat  produced  is  a  minimum.  Show  that 
the  currents  Ih  1 '2,  h  in  the  three  wires  will  satisfy  the  equations 

I\.Rv  =  I2R2  =  I3R3. 

4.  A  particle  attracted  toward  each  of  three  points  A,  B,  C  with  a 
*"force  proportional  to  the  distance  will  be  in  equilibrium  when  the  sum 


Chap.  XI.  PARTIAL  DIFFERENTIATION  139 

of  the  squares  of  the  distances  from  the  points  is  least.     Find  the  posi- 
|  tion  of  equilibrium. 

aHj:  Show  that  the  triangle  of  greatest  area  with  a  given  perimeter  is 
equilateral. 

6.  Two  adjacent  sides  of  a  room  are  plane  mirrors.  A  ray  of  light 
starting  at  P  strikes  one  of  the  mirrors  at  Q,  is  reflected  to  a  point  R  on 
the  second  mirror,  and  ic  there  reflected  to  S.  If  P  and  S  are  in  the 
same  horizontal  plane  find  the  positions  of  Q  and  R  so  that  the  path 
PQRS  may  be  as  short  as  possible. 

7.  A  table  has  four  legs  attached  to  the  top  at  the  corners  Ai,  Ai, 
As,  Aa  of  a  square.  A  weight  W  placed  upon  the  table  at  a  point  of  the 
diagonal  AiAs,  two-thirds  of  the  way  from  Ax  to  A3,  will  cause  the  legs 
to  shorten  the  amounts  «i,  s2,  S3,  84,  while  the  weight  itself  sinks  a  dis- 
tance h.  The  increase  in  potential  energy  due  to  the  contraction  of  a 
leg  is  ks2,  where  k  is  constant  and  s  the  contraction.  The  decrease  in 
potential  energy  due  to  the  sinking  of  the  weight  is  Wh.  The  whole 
system  will  settle  to  a  position  such  that  the  potential  energy  is  a  mini- 
mum. Assuming  that  the  top  of  the  table  remains  plane,  find  the 
ratios  of  4,  $2,  S3,  54. 


SUPPLEMENTARY  EXERCISES 

CHAPTER  III 

Find  the  differentials  of  the  following  functions: 

.      Vox2  +  b  , 

!•    ^ •  6.   x  (a2  +  x2)  Va?  -  x2. 

2     x  7     (2x  +  l)(2x  +  7)2 

b  Vax2  +  6  (2x  +  5)3 

3  2Va^±bxt  8     (a;  +  2)6  (a;  +  4)2 

6x  '  '    (x  +  l)2  (x  +  3)6' 

4     2ax  +  6       .  (2x2-l)  VxH7! 

Va.x2  +  5x  +  c  y-  x3 

(ax  +  fr)n+2     bjax+b)^1  n-i 

5'     a2  (n  +  2)        a2  (n  +  1)   '  10-  x  (*"  +  n>    n  ' 

Find  -^  in  each  of  the  following  cases: 

11.  2x2-  4:xy  +  Sy2  =  6a:  -4?/ +  18. 

12.  x3  +  3  x2z/  =  ?/3. 

13.  x  =  3y2  +  2  y\ 

14.  (x2  +  ?/2)2  =  2a2(x2  -y2). 

15.  x  =  Z  +  ; t,    y  =  2t- 


t-V    y  (J--1) 

*  1 

16.   x  =     ,     ,      >      ?/  = 


2 


Vl   +  f*  *    '  "    Vl    -f2 

17.  x  =  *  (J2  +  a2)*    «/  =  t  (t2  +  a2)f. 

18.  x  =  z2  +  2  s,        2  =  i/2  +  2  ?/. 

19.  x2  +  z2  =  a2,      2/2  =  62. 

20.  The  volume  elasticity  of  a  fluid  is  e  =  —  v  -j- .     If  a  gas  expands 

according  to  Boyle's  law,  pv  =  constant,  show  that  e  =  p. 

21.  When  a  gas  expands  without  receiving  or  giving  out  heat,  the 
pressure,  volume,  and  temperature  satisfy  the  equations 

pv  =  RT,    pvn  =  C, 
ft,  n,  and  C  being  constants.    Find  -pp  and  -r=- 

140 


SUPPLEMENTARY  EXERCISES  141 

22.  If  v  is  the  volume  of  a  spherical  segment  of  altitude  h,  show  that 
-Tj-  is  equal  to  the  area  of  the  circle  forming  the  plane  face  of  the  segment. 

23.  If  a  polynomial  equation 

/  (x)  -  0 
has  two  roots  equal  to  r,  /  (x)  has  (x  —  r)2  as  a  factor,  that  is, 

/  (x)  =  (x  -  r)»/i  (*), 
where  /i  (x)  is  a  polynomial  in  x.     Hence  show  that  r  is  a  root  of 

r  (x)  =  o, 

where/'  (x)  is  the  derivative  of/  (x). 

Show  by  the  method  of  Ex.  23  that  each  of  the  following  equations 
has  a  double  root  and  find  it: 

24.  x3  -  3  x2  +  4  =  0. 

25.  x3  -  x2  -  5  x  -  3  =  0. 

26.  4x3-8x2-3x  +  9  =  0. 

27.  4  x4  -  12  x3  +  x2  +  12  3  +  4  =  0. 

Find  -7-  and  -j-r  in  each  of  the  following  cases. 
ax  ax2 

28.  y  -  x  Va2  -  x2.  31.   ax  +  6y  +  c  =  0. 

X2  32.   x  =  2  +  3  f,  y  =  4  -  5 1 . 

^"-fc+iF  33.  *  =  -i,,  f-  <2 


30.   XT/  =  a2.  f  +  ! '  *  +  ! 

34.  If  „  =  *,  find  g  and  g. 

35.  Given  x2  —  y2  =  1,  verify  that 

dfy  dtxfdy\*t 

dx2  dy2  \dxj 

36.  If  n  is  a  positive  integer,  show  that 

-j-^  xn  =  constant. 

37.  If  u  and  v  are  functions  of  x,  show  that 

d4       .       dlu  dzu     dv  (Pu     drv  du      dh  d*v 

dr{UV)  "  dx~*"V  +     dx~>  '  Tx  +     dx~2  '  dx2_h dx  '  dx*~tUdx*' 

Compare  this  with  the  binomial  expansion  for  (u  +  v)4. 

38.  If/  (x)  =  (x  —  r)3/i  (x),  where /1  (x)  is  a  polynomial,  show  that 

f  (r)  =  /"  (r)  =  0. 


142  DIFFERENTIAL  CALCULUS 

CHAPTER   IV  I 

39.  A  particle  moves  along  a  straight  line  the  distance 

s  =  4/3  -  21/2  +  36£  +  1 
feet  in  t  seconds.     Find  its  velocity  and  acceleration.     When  is  the 
particle  moving  forward?     When  backward?     When  is  the  velocity 
increasing?     When  decreasing? 

40.  Two  trains  start  from  different  points  and  move  along  the  same 
track  in  the  same  direction.  If  the  train  in  front  moves  a  distance  6  t3 
in  t  hours  and  the  rear  one  12  t2,  how  fast  will  they  be  approaching  or 
separating  at  the  end  of  one  hour?  At  the  end  of  two  hours?  When 
will  they  be  closest  together? 

41.  If  s  =  V£,  show  that  the  acceleration  is  negative  and  propor- 
tional to  the  cube  of  the  velocity. 

42.  The  velocity  of  a  particle  moving  along  a  straight  line  is 

v  =  2 t2  -  3 1. 

Find  its  acceleration  when  t  =  2. 

k 

43.  If  v2  =  -,  where  k  is  constant,  find  the  acceleration. 

s 

44.  Two  wheels,  diameters  3  and  5  ft.,  are  connected  by  a  belt. 
What  is  the  ratio  of  their  angular  velocities  and  which  is  greater? 
What  is  the  ratio  of  their  angular  accelerations? 

45.  Find  the  angular  velocity  of  the  earth  about  its  axis  assuming 
that  there  are  365 1  days  in  a  year. 

46.  A  wheel  rolls  down  an  inclined  plane,  its  center  moving  the 
distance  s  =  5  t2  in  t  seconds.  Show  that  the  acceleration  of  the 
wheel  about  its  axis  is  constant. 

47.  An  amount  of  money  is  drawing  interest  at  6  per  cent.  If  the 
interest  is  immediately  added  to  the  principal,  what  is  the  rate  of. 
change  of  the  principal? 

48.  If  water  flows  from  a  conical  funnel  at  a  rate  proportional  to 
the  square  root  of  the  depth,  at  what  rate  does  the  depth  change? 

49.  A  kite  is  300  ft.  high  and  there  are  300  ft.  of  cord  out.  If  the 
kite  moves  horizontally  at  the  rate  of  5  miles  an  hour  directly  away 
from  the  person  flying  it,  how  fast  is  the  cord  being  paid  out? 

50.  A  particle  moves  along  the  parabola 

100?/  =  16  z2 
in  such  a  way  that  its  abscissa  changes  at  the  rate  of  10  ft./  sec.    Find 
the  velocity  and  acceleration  of  its  projection  on  the  ?y-axis. 

51.  The  side  of  an  equilateral  triangle  is  increasing  at  the  rate  of 
10  ft.  per  minute  and  its  area  at  the  rate  of  100  sq.  ft.  per  minute. 
How  large  is  the  triangle? 


SUPPLEMENTARY  EXERCISES  143 

CHAPTER   V 

52.   The  velocity  of  waves  of  length  \  in  deep  water  is  proportional  to 


\i 


X      a 

a      X 


when  a  is  a  constant.     Show  that  the  velocity  is  a  minimum  when 
X  =  a. 

53.  The  sum  of  the  surfaces  of  a  sphere  and  cube  is  given.  Show 
that  the  sum  of  the  volumes  is  least  when  the  diameter  of  the  sphere 
equals  the  edge  of  the  cube. 

54.  A  box  is  to  be  made  out  of  a  piece  of  cardboard,  6  inches  square, 
by  cutting  equal  squares  from  the  corners  and  turning  up  the  sides. 
Find  the  dimensions  of  the  largest  box  that  can  be  made  in  this  way. 

55.  A  gutter  of  trapezoidal  section  is  made  by  joining  3  pieces  of 
material  each  4  inches  wide,  the  middle  one  being  horizontal.  How 
wide  should  the  gutter  be  at  the  top  to  have  the  maximum  capacity? 

56.  A  gutter  of  rectangular  section  is  to  be  made  by  bending  into 
shape  a  strip  of  copper.  Show  that  the  capacity  of  the  gutter  will  be 
greatest  if  its  width  is  twice  its  depth. 

57.  If  the  top  and  bottom  margins  of  a  printed  page  are  each  of 
width  a,  the  side  margins  of  width  b,  and  the  text  covers  an  area  c, 
what  should  be  the  dimensions  of  the  page  to  use  the  least  paper? 

58.  Find  the  dimensions  of  the  largest  cone  that  can  be  inscribed 
in  a  sphere  of  radius  a. 

59.  Find  the  dimensions  of  the  smallest  cone  that  can  contain  a 
sphere  of  radius  a. 

60.  To  reduce  the  friction  of  a  liquid  against  the  walls  of  a  channel, 
the  channel  should  be  so  designed  that  the  area  of  wetted  surface  is  as 
small  as  possible.  Show  that  the  best  form  for  an  open  rectangular 
channel  with  given  cross  section  is  that  in  which  the  width  equals 
twice  the  depth. 

61.  Find  the  dimensions  of  the  best  trapezoidal  channel,  the  banks 
making  an  angle  6  with  the  vertical. 

62.  Find  the  least  area  of  canvas  that  can  be  used  to  make  a  conical 
tent  of  1000  cu.  ft.  capacity. 

63.  Find  the  maximum  capacity  of  a  conical  tent  made  of  100  sq.  ft. 
of  canvas. 

64.  Find  the  height  of  a  light  above  the  center  of  a  table  of  radius  a, 
so  as  best  to  illuminate  a  point  at  the  edge  of  the  table;  assuming  that 
the  illumination  varies  inversely  as  the  square  of  the  distance  from  the 
light  and  directly  as  the  sine  of  the  angle  between  the  rays  and  the 
surface  of  the  table. 


144  DIFFERENTIAL  CALCULUS 

65.  A  weight  of  100  lbs.,  hanging  2  ft.  from  one  end  of  a  lever,  is  to 
be  raised  by  an  upward  force  applied  at  the  other  end.  If  the  lever 
weighs  3  lbs.  to  the  foot,  find  its  length  so  that  the  force  may  be  a 
minimum. 

66.  A  vertical  telegraph  pole  at  a  bend  in  the  line  is  to  be  supported 
from  tipping  over  by  a  stay  40  ft.  long  fastened  to  the  pole  and  to  a 
stake  in  the  ground.  How  far  from  the  pole  should  the  stake  be 
driven  to  make  the  tension  in  the  stay  as  small  as  possible? 

67.  The  lower  corner  of  a  leaf  of  a  book  is  folded  over  so  as  just  to 
reach  the  inner  edge  of  the  page.  If  the  width  of  the  page  is  6  inches, 
find  the  width  of  the  part  folded  over  when  the  length  of  the  crease  is 
a  minimum. 

68.  If  the  cost  of  fuel  for  running  a  train  is  proportional  to  the 
square  of  the  speed  and  $10  per  hour  for  a  speed  of  12  mi./hr.,  and 
the  fixed  charges  on  $90  per  hour,  find  the  most  economical  speed. 

69.  If  the  cost  of  fuel  for  running  a  steamboat  is  proportional  to 
the  cube  of  the  speed  and  $10  per  hour  for  a  speed  of  10  mi./hr.,  and 
the  fixed  charges  are  $14  per  hour,  find  the  most  economical  speed 
against  a  current  of  2  mi./hr. 


CHAPTER  VI 

Differentiate  the  following  functions: 

76.  sec2  x  —  tan2  x. 

77.  sin3  -  sec  o* 
x       3 

78.  tan 


70. 

sin  re 

X 

71. 

sin0 

1  —  cos  0 

72. 

1  +  cos  0 

sin0 

73. 

sin  ax  cos  ax 

79. 


1  -x 
2  tanz 
1  —  tan2  x 


L  e  0  80.   5  sec7  0  -  7  sec6  0. 

74.  cot 7:  —  esc-*  01  n     , 

2  2  81.   sec  x  esc  x  —  2  cot  x. 

75.  tan  2  x  —  cot  2  x. 

Differentiate  both  sides  of  each  of  the  following  equations  and  show 
that  the  resulting  derivatives  are  equal. 

82.  sec2  x  +  csc2£  =  sec2  x  esc2  x. 

83.  sin2z  =  2  sin  £  cos  z. 

84.  sin  3  x  =  3  sin  x  —  4  sin3  x. 

85.  sin  (x  +  a)  =  sin  x  cos  a  +  cos  x  sin  a. 

86.  sec2  x  =  1  +  tan2  x. 

87.  sin  x  +  sin  a  =  2  sin  §  {x  -f-  a)  cos  \{x  —  a). 


SUPPLEMENTARY  EXERCISES  145 

88.  cos  a  -  cos  x  =  2  sin  h  (x  +  a)  sin  \  (x  —  a). 

Find  -~  and  -rK  in  each  of  the  following  cases: 
ax  dx- 

89.  x  =  a  cos2  6,  y  =  a  sin2  0. 

90.  x  =  a  cos5  0,  y  =  a  sin5  0. 

91.  x  =  tan  0  —  d,        y  =  cos0. 

92.  x  =  sec2  0,  y  =  tan2  0. 

93.  x  =  sec0,  ?/  =  tan0. 

94.  x  =  esc  0  —  cot  0,  y  =  esc  0  +  cot  0. 

Differentiate  the  following  functions: 

95.  sin-  \J\-  102-    «  csc-'  I  +  *>=*• 

96.  008-  gV  103-    if?-"*** 

-2x\  104.    Vl  —  x  sin-1  x  —  Vx- 

105.    sec-1 r  +  sin-1  — —  • 

2  2s  +  1  x  —  1  x  +  1 


a11"  i1^} 


97.  t 

2 

98.  — ^  tan"1 


V3  V3  in,       •  _'c+6cosx 

v  °  °  106.    sin  l  t — ; 

b  +  a  cos  x 


.T 


99.    cos     Vi^TI  107.    iUos-lx  +  ^vT^. 

V5  

100.    csc  x2x_x'  108.     Vf-a2  -asec"1-- 


101.    sec 


-ihl} 


a 


109.   e^5.  118.    Itan-^+hn^  +  z2)' 


110.    Ve*. 


121.   (*  +  I)]n(*  +  i)-s-i. 


a  a      2 

119.   e-«cos(o+60. 

}£  Linl  12°-   «n(a  +  V^+^). 

1 

113.  7X. 

114.  a*  In  x.  \rx  +  a  +  Vx  -  a 

115.  In  sinn  x.  \/x  +  a  -  y/x  -  a 

116.  lnln.r.  123.   tan"1  J  (e*  +  e"«). 

117.  lnf^^Y  124.   ln(Vx  + Vx^F2). 

125.  (3  +  1)  In  (:r2  +  2  x  +  5)  +  |  tan"*  £+_. 

126.  sec  £  a;  tan  \  x  +  In  (sec  \x  +  tan  |  x). 


146  DIFFERENTIAL  CALCULUS 

127.  x  sec"1  Ux  +  ij  -  In  (x2  +  1) . 

128.  | In  (|z2  +  lU|z  +  tan-i|a;. 


CHAPTER  VII 

Find  the  equations  of  the  tangent  and  normal  to  each  of  the  follow- 
ing curves  at  the  point  indicated: 

129.  y2  =  2  x  +  y,  at  (1,  2). 

130.  x2  -  y2  =  5,  at  (3,  2). 

131.  z2  +  2/2  =  x  +  3y,  at  (-1,  1). 

132.  x*  +  y*  =  2,  at  (1,  1). 

133.  y  =  lnz,  at  (1,  0). 

134.  x2  ix  +  y)  =  a2  (x  -  y),  at  (0,  0). 

135.  x  =  2  cos  6,  y  =  3  sin  0,  at  0  =  -• 

136.  r  =  a  (1  +  cos0),  at  0  =  ~ 

Find  the  angles  at  which  the  following  pairs  of  curves  intersect: 

137.  x2  +  y2  =  8  x,  y2  (2  -  x)  =  xK 

138.  y2  =  2ax  +  a2,  x2  =  2  by  +  62. 

139.  x2  =  lay,  (x2  +  4a2)y  =  8a3. 

140.  y2  =  6z,  x2  +  i/2  =  16. 

141.  p  =  |  (e*  +  <r*  ),  y  =  1. 

142.  y  =  sinx,  y  =  sin  2  re. 

143.  Show  that  all  the  curves  obtained  by  giving  different  values  to 
n  in  the  equation 

are  tangent  at  (a,  b). 

144.  Show  that  for  all  values  of  a  and  b  the  curves 

xz  —  3  z?/2  =  a,     3  x2y  —  y3  =  b, 
intersect  at  right  angles. 

Examine  each  of  the  following  curves  for  direction  of  curvature  and 
points  of  inflection: 

1AK  1    —  X 

145'   »  ■  T  +  rf- 

146.  y  =  tan  x. 

147.  x  =  6  2/2  -  2  2/3. 


SUPPLEMENTARY  EXERCISES  147 

148.  x  =  2*-i,  y  =  2t  +  ±- 

149.  Clausius's  equation  connecting  the  pressure,  volume,  and  tem- 
perature of  a  gas  is 

RT  c 

P==v-a       Ttv  +  b)3' 
R,  a,  b,  c  being  constants.     If  T  is  constant  and  p,  v  the  coordinates 
of  a  point,  this  equation  represents  an  isothermal.     Find  the  value  of 
T  for  which  the  tangent  at  the  point  of  inflection  is  horizontal. 

150.  If  two  curves  y  =  /  (.c),  y  =  F  (x)  intersect  at  x  =  a,  and 
/'  (a)  =  F'  (a),  but  /"  (a)  is  not  equal  to  F"  (a),  show  that  the  curves 
are  tangent  and  do  not  cross  at  x  =  a.  Apply  to  the  curves  y  =  x2 
and  y  =  x3  at  x  =  0. 

151.  If  two  curves  y  =  f  (x),  y  =  F  (x)  intersect  at  x  =  a,  and 
f  (a)  =  F'  (a),  /"  (a)  =  F"  (a),  but  /'"  (a)  is  not  equal  to  F'"  (a), 
show  that  the  curves  are  tangent  and  cross  at  x  =  a.  Apply  to  the 
curves  y  =  x-  and  y  =  x2  +  (x  —  l)3  at  x  =  1. 

Find  the  radius  of  curvature  on  each  of  the  following  curves  at  the 
point  indicated: 

152.  Parabola  y2  =  ax  at  its  vertex. 

x2      v2 

153.  Ellipse  —„  +  rr  =  1  at  its  vertices. 

a2      o' 

154.  Hyperbola  ^-f£  =  lata;=  Va2  +  b2. 

a2      o2 

155.  y  =  In  esc  x,  at  (  ^ ,  0  j  • 

156.  x  =  §  sin  y  —  ^  In  (sec  ?/  +  tan  y),  at  any  point  (x,  y). 

157.  x  =  acos30,  y  =  asin30,  at  any  point. 

158.  Find  the  center  of  curvature  of  y  =  In  (x  —  2)  at  (3,  0). 

Find  the  angle  \p  at  the  point  indicated  on  each  of  the  following 
curves: 

159.  r  =  20,  at  d  =  0. 

160.  r  =  a  +  b  cos  0,  at  0  =  ^« 

161.  r  (1  —  cos0)  =  fc,  at  0  =  ^« 

162.  r  =  a  sin  2  0,  at  0  =  v 

4 

Find  the  angles  at  which  the  following  pairs  of  curves  intersect: 

163.  r  (1  —  cos0)  =  a,     r  =  a  (1  —  cos0). 

164.  r  =  a  sec2  x,  r  =  b  esc2  =• 

-  J 


148  DIFFERENTIAL  CALCULUS 

165.  r  =  a  cos  d,  r  =  a  cos  2  0. 

166.  r  =  a  sec  0,  r  =  2  o  sin  0. 

Find  the  equations  of  the  tangent  lines  to  the  following  curves  at 
the  points  indicated: 

o 

167.  x  =  2 1,  y  =  - ,  z  =  t2,  at  t  =  2. 

168.  x  =  sint,  y  =  cost,  z  =  sec/,  at  t  =  0. 

169.  x2  +  y2  +  22  =  6,  x  +  y  +  z  =  2,  at  (1,  2,  -1). 

170.  2  =  x2  +  y\  z2  =  2x  -  2y,  at  (1,  -1,  2). 


CHAPTER  VIII 

171.  A  point  describes  a  circle  with  constant  speed.  Show  that 
its  projection  on  a  fixed  diameter  moves  with  a  speed  proportional 
to  the  distance  of  the  point  from  that  diameter. 

172.  The  motion  of  a  point  (x,  y)  is  given  by  the  equations 

x  =  7=  V  a2  —  t2  +  -7T  sin  x  - , 
2  2  a 

2/  =  |  V^+T2  +f  In  (l  +  VoM=T2)- 

Show  that  its  speed  is  constant. 

Find  the  speed,  velocity,  and  acceleration  in  each  of  the  following 
cases: 

173.  x  =  2  +  St,  y  =  4-9*. 

174.  x  =  a  cos  {(at  +  a),  y  =  a  sin  (cot  +  a). 

175.  x  =  a  -\-  at,  y  =  b  +  /3t,  z  =  c  +.  yt. 

176.  x  =  el  sin  t,    y  =  el  cos  t,  z  =  kt. 

177.  The  motion  of  a  point  P  (a;,  ?/)  is  determined  by  the  equations 

x  =  a  cos  (nt  +  a),  y  =  b  sin  (n£  +  a). 

Show  that  its  acceleration  is  directed  toward  the  origin  and  has  a 
magnitude  proportional  to  the  distance  from  the  origin. 

178.  A  particle  moves  with  constant  acceleration  along  the  parab- 
ola y2  =  2  ex.     Show  that  the  acceleration  is  parallel  to  the  x-axis. 

179.  A  particle  moves  with  acceleration  [a,  o]  along  the  parabola 
y2  =  2  ex.     Find  its  velocity. 

180.  Show  that  the  vector  I  j-j  >  j~f  I  extends  along  the  normal  at 
{x,  y)  and  is  in  magnitude  equal  to  the  curvature  at  (x,  y). 


SUPPLEMENTARY  EXERCISES  149 

CHAPTER   IX 

181.  Show  that  the  function 

x*  -  1 

vanishes  at  x  =  —  1  and  x  =  1,  but  that  its  derivative  does  not  vanish 
between  these  values.     Is  this  an  exception  to  Rolle's  theorem? 

182.  Show  that  the  equation 

^-5x4-4  =  0 

has  only  two  distinct  real  roots. 

183.  Show  that 

x2  sin  - 

Lim  — : =  0, 

x  =  o    sin  x 

but  that  this  value  cannot  be  found  by  the  methods  of  Art.  73.     Explain. 

184.  Show  that 

T  .      1  —  cos  x 

Lim  =  0. 

x=o      cos  a: 

Why  cannot  this  result  be  obtained  by  the  methods  of  Art.  73? 
Find  the  values  of  the  following  limits: 

185.  Lim    **'  ~  »  .  189.    Lim  (I  -  2**Y 

x  =  0  1   -  COS2X  x  =  0\X2  X      J 

,00     t-      ^3~r  -    Vl2  -x  190.    LimxV-*2. 

186.  Lim 


x=^2x  -  3  Vl9  -5 


x 


ttx  191.   Lim  - 


x  =  oo 

xln  x 


tan  -jr  x  =  o  sin2  z  —  *  cot  * 

187.  Lim-— -r 77-  T.  ± 

x  =  i  1  +  csc  (x  —  1)  192.   Lim  (,sec  x)  *2- 

188.  Lim'"('-f. 

z  i  1       «Ot   (7TX) 

193.  The  area  of  a  regular  polygon  of  n  sides  inscribed  in  a  circle 
of  radius  a  is 

„      .       7T  7T 

wa2sin  -  cos-- 
n        n 

Show  that  this  approaches  the  area  of  the  circle  when  n  increases 
indefinitely. 

194.  Show  that  the  curve 

x3  +  y3  =  3  xy 
is  tangent  to  both  coordinate  axes  at  the  origin. 


150  DIFFERENTIAL  CALCULUS 


CHAPTER  X 

Determine  the  values  of  the  following  functions  correct  to  four 
decimals : 

195.  cos  62°.  198.  ^U. 

196.  sin  33°.  199>   tan"1  (TV)- 

197.  In  (1.2).  200.  .esc  (31°). 

201.  Calculate  t  by  expanding  tan-1  x  and  using  the  formula 

\  =  tan-*  (1). 

202.  Given  In  5  =  1.6094,  calculate  In  24. 

203.  Prove  that 

D  =  Vfh 

is  an  approximate  formula  for  the  distance  of  the  horizon,  D  being  the 
distance  in  miles  and  h  the  altitude  of  the  observer  in  feet. 

Prove  the  following  expansions  indicating  if  possible  the  values  of 
x  for  which  they  converge : 

204.  In  (1  +  a?)  =  In  10  +  f  (x  -  3)  -  &  (x  -  3)2  +  •  •  •  .     . 

/v.2  /Y>4  syS 

V'205.    ln(e*+<r*)  =|_^+g+   .... 

206.  In  (1  +  sin  x)  =a;-|x2+|x3-T12i4+  •  •  •  . 

207.  ex  sec  x  =  1  +  x  +  x2  +  f  r5  +  hxi  +  •  •  •  . 

208.  In  (*  +  VI+T2)  =  *  -  if  +|r|f  +  •  •  •  . 

209.  in i±4-2g  +  i+^  +  ...]. 

210.  In  tan  x  =  In  x  +  -~  +  ^-pr-  +  •  •  •  . 
211.^*  =  l+x+|I-^- . 

212.  etan*  =  1  +*  +|  +  |f  +  ^47  +  *  '  '  ' 

Determine  the  values  of  x  for  which  the  following  series  converge: 

/y»2  /v»3  /y»4 

213.  1+35  +  1+1+^+.... 

214.  (s  -  1)  H ^i ' 33 1 4I h  •  •  •  . 

215.  1  -\-2x  +  3x2  +  4.x3+  •  •  •  . 
91A     9      *  +  2   ,    (s  +  2)«   ,    (a;  +  2)3 

216-   2  +  T^2■+-2T3-  +  -3T4~+,••• 


SUPPLEMENTARY  EXERCISES  151 


CHAPTER  XI 

In  each  of  the  following  exercises  show  that  the  partial  derivatives 
satisfy  the  equation  given: 

~.-  o  ,  du   .      du         du 

21/.    U=xy  +  y-Z;        ,_+,--„_ 

218.   z  =  x*  -  2x2if  +  if,      y%+xfy=°- 


219.   u  =  (x  + 


,  .  fdll        dll\  dU 

y)\nxz,      x[---)  =  z~. 
220.   „-(*+I)tan-i(„-i),      £  +  * 


du  du 

dy      y     dz 


221.   u  =  xy  +  -, 


dht,  _d*u^        %&u        8a*M 


222.  z  =  In  («"  +  t),     ||  +  ^  -  * 

223.  «  =  K±2, 

d2tt  d2u        a2M  _  &ut 

dxdy       dydz       dz2  "  dx2' 

Prove  the  following  relations  assuming  that  z  is  a  function  of  x  and  y\ 

224.  u=*(x  +  y-  z)2, 

du       du       du  d£        dudz 
dy      dx      dy  dx      dx  dy 

225.  u  =  z  +  ex!/, 

du           du  dz  dz 

X — ■  —  V =  X  —   —  V  — • 

dx      J  dy         dx      y  dy 

226.  u  =  z  {x2  -  y2), 

du   .       lu       ,.        ~  /    dz  dz  \ 

227.  If  x  =  |  (e$  +  e'"),     y  =  g  (e9  -  e-9),  show  that 

i¥)  -(£V 

\dr/0      \d£/y 

228.  If  xyz  =  a3,  show  that 

(§y\  (*l\  (qx\  m    « 

\dx)z\dylx\dz}y 


152 


DIFFERENTIAL  CALCULUS 


In  each  of  the  following  exercises  find  Az  and  its  principal  part, 
assuming  that  x  and  y  are  the  independent  variables.  When  Ax  and 
Ay  approach  zero,  show  that  the  difference  of  Az  and  its  principal  part 
is  an  infinitesimal  of  higher  order  than  Ax  and  Ay. 

229.  z  =  xy.  232.   z  =  Vx2  +  y\ 

230.  z  =  x2  -  y2  +  2  x. 

y 


231.   z  = 


x2  +  1 


Find  the  total  differentials  of  the  following  functions: 

233.  ax*  +  bx2y2  +  cy\ 

234.  In  O2  +  y2  +  z2). 

235.  x2  tan"1  ^  -  y2  tan"1  -• 

2/ 


re 


236.   yzex  +  z£ey  +  z?/e2 


U 


237.  If  w  =  xnf  0),     g  =  *,  show  that 

x 

du    ,        dU 

x  —  +  2/  — ■  =  nu. 
dx  ay 

238.  If  u  =  /  (r,  s),    r  =  a;  +  ?/,     s  =  x  —  y,  show  that 

ax     a^/  ~    a7" 

239.  liu  =f(r,s,t),     r  =  -,     s  =  ^,     «  =  -,  show  that 

?/  Z  X 

du    ,         dU     ,         dU         „ 

x- — \-y  - — V  z  — -  =  0. 
dx      u  dy  dz 

240.  If  a  is  the  angle  between  the  a>axis  and  the  line  OP  from  the 
origin  to  P  (x,  y,  z),  find  the  derivatives  of  a  in  the  directions  parallel 
to  the  coordinate  axes. 

241.  Show  that 

(cot  y  —  y  sec  x  tan  x)  dx  —  (x  esc2  y  +  sec  x)  dy 
is  an  exact  differential. 

Find  the  equations  of  the  normal  and  tangent  plane  to  each  of  the 
following  surfaces  at  the  point  indicated: 

242.  x2  +  2y2  -  z2  =  16,   at  (3,  2,  -1). 

243.  2  x  +  3  y  -  4  z  =  4,  at  (1,  2,  1). 

244.  z2  =  8xy,  at  (2,  1,  -4). 

245.  y  =  z2  -  x2  +  1,  at  (3,  1,  -3). 

246.  Show  that  the  largest  rectangular  parallel  opiped  with  a  given 
surface  is  a  cube. 


SUPPLEMENTARY  EXERCISES  153 

247.  An  open  rectangular  box  is  to  be  constructed  of  a  given  amount 
of  material.     Find  the  dimensions  if  the  capacity  is  a  maximum. 

248.  A  body  has  the  shape  of  a  hollow  cylinder  with  conical  ends. 
Find  the  dimensions  of  the  largest  body  that  can  be  constructed  from 
a  given  amount  of  material. 

249.  Find  the  volume  of  the  largest  rectangular  parallelopiped  that 
can  be  inscribed  in  the  ellipsoid. 

a2  ^  b2  ^  c2 

250.  Show  that  the  triangle  of  greatest  area  inscribed  in  a  given 
circle  is  equilateral. 

251.  Find  the  point  so  situated  that  the  sum  of  its  distances  from 
the  vertices  of  an  acute  angled  triangle  is  a  minimum. 

252.  At  the  point  (x,  y,  z)  of  space  find  the  direction  along  which  a 
given  function  F  (x,  y,  z)  has  the  largest  directional  derivative. 


ANSWERS   TO  EXERCISES 

Page  8 

1.  -f.  4.    -1. 

2.  V2.  5.    1. 

O.      ~ "X.  O.      2' 

Page  14 
*3.   2. 

5.   The  tangents  are  parallel  to  the  z-axis  at  (  —  1,  —1),  (0,  0),  and 

(1,  —1).     The  slope  is  positive  between  (  —  1,  —1)  and  (0,  0) 

and  on  the  right  of  (1,  —1). 

10.  Negative. 

11.  Positive  in  1st  and  4th  quadrants,  negative  in  2nd  and  3rd. 

Pages  27,  28 

31.  When  x  =  4,  y  =  f  and  dy  =  0.072  dx.     When  x  changes  to  4.2, 

dx  =  0.2  and  an  approximate  value  for  y  is  y  +  dy  =  0.814. 
This  agrees  to  3  decimals  with  the  exact  value. 

32.  When  x  =  0,  the  function  is  equal  to  1  and  its  differential  is  —  dx. 

When  x  =  0.3,   an   approximate  value  is  then   1  —  dx  =  0.7. 
The  exact  value  is  0.754. 
34.    18.  35.    (a,  2  a). 

36.   Increases  when  x  <  x,  decreases  when  x  >  =■ 

o  o 

39         2     _     (y-w 

d9'        {x  -  1)2  ~  2 


Page  31 

x  a2 


1 

37. 

*»=fc£2- 

38. 

tan-1 1. 

1. 

2 

(z-1)2' 

3. 

(x  -  l)2  (x  4 

2         4 

4. 

y'       y3' 

x         a2 

5. 

.y>             yZ 

6. 

X                     1 

2^'        42Z3 

2.    - 


(z  ~  I)3  Va2  -^         (a2  __  X2)% 

-  2)3  (7s  +  2),     (a;  -  •  1)  (x  +  2)2  (42 z2  +  24z  -  12). 

1-2/  2 


7. 


re  —  1 '     (x  —  1) 


3 


8.    -^i       a§ 


3*      3  X*  y* 


154 


ANSWERS  TO  EXERCISES  155 


ffiy  12  ffix  12 

l6'    dx*''        /(2  +  303'    dif"t(2-St)3' 


Pages  36-38 

1.  v  =  100  -  32 1,  a  -  -  32.     Rises    until  <  =  3}.     Highest   point 

h  =  206.25. 

2.  v  =  /3  -  12  r-  +  32  f,    a  =  3  /2  -  24  <  +  32.     Velocity    decreasing 

between  £  =  1.691  and  t  =  6.309.     Moving  backward  when  t 
is  negative  or  between  4  and  8. 

7.   u  =  fr  —  2  c/,  a:  =  —  2  c.     Wheel  comes  to  rest  when  t  =  —• 

£  c 

10.  9tt  ru.  ft./min.  15.    12*  ft./sec,     7£  ft./sec. 

11.  144  7T  sq.  ft./sec.  ,  16.   4  V3  mi./hr. 

12.  Decreasing  8  7r  cu.  ft./sec.  ctan/3 

Ai     1  17.    ir-  ft./sec. 

13.  v3  :  1.  Tra2 

14.  \  \x3  in. /sec. 

■8.    Neither  approaching  nor  separating. 
19.   25.8  ft./sec.  20.   64  V3  ft./sec. 


Pages  43-45 


rl.  Minimum  3$.  2.    Minimum —10,  maximum  22. 

3.  Maximum  at  x  =  0,  minima  at  .t  =   —1  and  x  =  -f  1. 

4.  Minimum  when  x  =  0.  13     2  a  V3. 
10.  f^2. 

14.  Length  of  base  equals  twice  the  depth  of  the  box. 

16.  Radius  of  base  equals  two-thirds  of  the  altitude. 

4 

17.  Altitude  equals  -  times  diameter  of  base. 

7T 

I671-V3  20.   Girth  equals  twice  length. 

18.    *  .  /— 

27  21.    Radius  equals  2  v  6  inches. 

19.  \  (aj  +  a2  +  a3  +  04). 

22.  The  distance  from  the  more  intense  source  is  ^2  times  the  dis- 

tance from  the  other  source. 

23.  12  V%  25.    19f  ft. 

24.  [5s  +  6s]5. 

26.  Radius  of  semicircle  equals  height  of  rectangle. 

27.  4  pieces  6  inches  long  and  2  pieces  2  ft.  long. 

28.  The  angle  of  the  sector  is  two  radians. 

29.  At  the  end  of  4  hours. 


156  DIFFERENTIAL  CALCULUS 

* 

30.  He  should  land  4.71  miles  from  his  destination. 

31.  — x— ,  a  being  the  length  of  side. 

35.   2|  mi./hr.  -  36.    13.6  knots. 


Page  48 

1.  Maximum  =  a,  minimum  =  —a. 

2.  Maximum  =  0,  minimum  =  —  (/y)  • 

3.  Minimum  =  —  1. 


3.  Minimum  =  —  1. 

4.  Minimum  =  0,  maximum  =  247. 

lO        TTifViQT.   A  nr>   K 


10.   Either  4  or  5. 

Pages  52,  53 
19.  A  =  3.  00  -k      ,    . 

20    A  =  -  7-     B  =  -  3  n*"      6'  n       ng  any  mteSer- 

21.   V3  -  f . 

23.  Velocity  =  — 2  7rn^l,  acceleration  =  0. 

24.  —^-  miles  per  minute.  26,   I  +  fjr  "^3. 

6  ,.  ,  28.    13V13. 

25.  |  radians  per  hour. 

29.  The  needle  will  be  inclined  to  the  horizontal  at  an  angle  of  about 

32°  30' 

30.  120°.  „2     a 

31.  120°.  '    7T* 

33.   If  the  spokes  are  extended  outward,  they  will  form  the  sides  of  an 
isosceles  triangle. 

Page  56 

v 

24.  03  =  -  cos  <f>,  r  being  the  radius  of  pulley  and  </>  the  angle  formed 

r 

by  the  string  and  line  along  which  its  end  moves. 

25.  4V35. 

Page  61 

27.  x  =  rnr  +  cot-1 2,  n  being  any  integer. 
30.  x  <  -3,    x  >  2,    or  -2  <  x  <  1. 

Pages  65,  66 

1.  2y  -x  =  5,     y  +  2x  =  0. 

2.  ?/  +  4z  =  8,     4?/  -re  =  15. 

S.  2y=F  x  =  ±a,    y±2x=±3a. 


ANSWERS  TO  EXERCISES  157 

4.  y  =  a  (xlnb  -\- 1),    x  +  ay  In  b  =  a2  In  b. 

s. , -|-1  v5(,-f),  »-i+|^(«-l)-a 

6.  2l  +  Sfi  «  i      fay  -  a'Ulx  =  (62  -  o»)  I*. 

7.  x  +  2/  =  2,    x  -  y  =  2.  12.  90°. 

8.  x  +  3y  =  4,    2/-3x  =  28.  13.  tan"1 2  V2. 

9.  y  +  x  tan  &  4i  =  a4>i  tan  §  <fr.  _  In  10  —  1 
10.  90°,     tan"1!.  14-  tan  'in  10  +  1* 

U*   45°'  15.  tan"1 3  V3. 


Page  70 

1.  Point  of  inflection  (0,  3).     Concave  upward  on  the  right  of  this 

point,  downward  on  the  left. 

2.  Point  of  inflection  (§,  —  f).     Concave  upward  on  the  right  of  thia 

point,  downward  on  the  left. 

3.  The  curve  is  everywhere  concave  upward.     There  is  no  point  of 

inflection. 

4.  Point  of  inflection  (1,  0).     Concave  on  the  left  of  this  point,  down- 

ward on  the  right. 

5.  Point  of  inflection!  —2,  — -5K     Concave  upward  on  the  right, 

downward  on  the  left. 

6.  Points  of  inflection  at  x  =  ±—7=..    Concave  downward  between 

V2 
the  points  of  inflection,  upward  outside. 

7.  Points  of  inflection   (0,  0),   (±3,  ±1).      Concave  upward  when 

-3  <x  <  Oor  x>3. 

8.  Point  of  inflection  at  the  origin.     Concave  upward  on  the  left  of 

the  origin. 


Pages  76,  77 

7.  *•• 

a 

8.  secy. 

(y2  + 1)2 


1. 

±2  VS. 

3. 

a2 

■  —  • 

b 

4. 

3V2. 

5. 

ir 

e1  Vi2' 

6. 

I  a. 

9. 


iy 


10.  2  a  sec3 ~ 


158  DIFFERENTIAL  CALCULUS 


Page  79 


There  are  two  angles  \p  depending  on  the  direction  in  which  s  is 
measured  along  the  curve.  In  the  following  answers  only  one 
of  these  angles  is  given. 


1.   tan"1 

2 


i) 


7T 
I" 


3'        l" 


4. 

tan"1  (-2). 

5. 

**■• 

7. 

0°,  90°,  and  tan"1  3  VB. 

8. 

6  =  ±|ir. 

10. 

O/l 

3. 

Page  84 


1. 


X 


V2  2        "      a  ■      t 

o.  tan  x— =« 
n     x  —  e       1                z  —  1  V2 

2.    =  1  —  ye  = 


3. 


s-e*  y  _2      2 

*  ?  k 

e2  e* 


2  7.   69°  29'. 

irk 


Pages  92,  93 

OLD 

1.  The  angular  speed  is  -z—. — -,  where  x  is  the  abscissa  of  the  moving 

a2  +  x2 

point. 

2.  If  #i  is  the  abscissa  of  the  end  in  the  rc-axis  and  yi  the  ordinate  of 

the  end  in  the  ?/-axis,  the  velocity  of  the  middle  point  is 


r  ,  x     _  vxi~\ 


2  2/i. 
the  upper  signs  being  used  if  the  end  in  the  #-axis  moves  to  the 


right,  the  lower  signs  if  it  moves  to  the  left.     The  speed  is 


av 


2yi 
3.   The  velocity  is 

[v  —  aoj  sin  6,     aoi  cos  0], 

where  6  is  the  angle  from  the  rc-axis  to  the  radius  through  the 
moving  point.     The  speed  is 

vV  -|-  a2co2  —  2  aw  sin  0. 

6.  The  boat  should  be  pointed  30°  up  the  river. 

7.  Velocity  =  [a,  b,  c  —  gt],  Acceleration  =  [0,  0,  —g]t 

Speed  =  Va2  +  62  +  (e  -  gt)2. 


ANSWERS  TO  EXERCISES  159 


9.  Velocity  =  [aa>  (1  —  cos0),     a<o  sin  0], 

Speed  =  aia  V2  —  2  cos  0  =2  aw  sin  \  0, 
Acceleration  =  [aco2  sin  0,     aa>2  cos  0]. 


|~      3  v2  sin  §  B      3  ?>2  cos  *  fl~j 
L      4a   sin  £0'     4  a   sin  \  d] 


12.  x  =  vt  cos  to/,     ?/  =  trt  sin  u>l.     The  velocity  is  the  sum  of  the  par- 

tial velocities,  but  the  acceleration  is  not. 

13.  x  =  a  cos  cot  +  b  cos  2  u>t}  y  =  a  sin  oot  -\-b  sin  2  a>/.     The   velocity  is 

the  sum  of  the  partial  velocities  and  the  acceleration  the  sum 
of  the  partial  accelerations. 

14.  x  =  aunt  —  a  sin  (an  +  a>2)  t,     y  =  a  cos  (coi  +  u2)  t.     The  velocity  is 

the  sum  of  the  partial  velocities  and  the  acceleration  the  sum 
of  the  partial  accelerations. 


A  &. 

V3.  n. 

.4.  0. 

/b.  ea. 

•6.  2. 

V7.  -2. 

t  8.  1. 

^9.  0. 

f.10.  7T2. 

vll.  1. 

12.  1. 

I  13.  0. 

1/14.  -h 

V\b.  0. 

K*6.  0. 

17.  -i 


1.  0.0872. 

V2.  0.8480. 

3.  1.0724. 

4.  1.6003. 
t/5.  1.0154. 


21.    (-2,1,0). 


Page  100 


18. 

i 

2« 

19. 

1. 

20. 
•^1. 

-3. 

a. 

22. 

1 

7T 

23. 

/'  (x)  dz. 

24. 

oo. 

25. 

i 

2- 

26. 

oo. 

27. 

GO. 

28. 

1. 

29. 

1. 

30. 

a. 

31. 

em 

Page  105 

6. 

0.1054. 

7. 

1.6487. 

8. 

0.0997. 

9. 

2.833. 

Page  118 

22. 

(1,  1,  2). 

160  DIFFERENTIAL  CALCULUS 

Pages  123,  124 

1.   Increment  =  —0.151,  principal  part  =  —0.154. 
.    dT      1  (dl      dg\      a.  ,    _  , 

~~T  =  2  w o/      felnce  rff  and  rf0  may  De  eitner  positive  or 

negative,  the  percentage  error  in  T  may  be  \  the  sum  of  the 

percentage  errors  in  /  and  g. 

5.   The  percentage  error  in  g  may  be  as  great  as  that  in  s  plus  twice 

that  in  T. 

iQ         u  +v  2  22  —  uy 

10. •  lo.    - • 

u  zx  —  2  uv 

14.    I  (x2  +  r/H-  xy  -  22). 
x 

Pages  131,  132 


2.  (i™)  =§L  +  ^L^. 

\dX/y       dX        dz  dx 


dx      dx      dy  dx 

,    '    dx  ""  dx      dydx      dz  [dx  dy  dx  '" 

dldF_dJ_dF  /13.3V3-4. 

_     d2       dy  dx       dx  dy  c 

dx  =       d/       d/df  14-    --3(^cosa+2/cos/8+2COS7). 

dy      d2  dy 

Page  135 

1.  ^p  =  S^  =  ^T^'     (x-l)+2(y-2)+2(2-2)=0. 

2.  Normal,  y  +  4  x  =  5,    2  =  3. 
Tangent  plane,  x  —  4y  —  14  =  0. 

x -  3      y  -  4      2-5 

3.  — 3~  =  ~~ 4- =  "ZJ  '     3x  +  4y-5z  =  0. 

.     x  —  5      y  —  1      2  —  3 

4.  —X~  =5— =  -33"'     x  +  5y-32-l=0. 

X-5t/-l2-l  c.rn 

5.  -— j-  =  — j—  =  — g— ,    x-4y-62  +  5  =  0. 

6.  x  +  z  =  y -z  =  ±  V2. 

Pages  138,  139 

^  1.  The  box  should  have  a  square  base  with  side  equal  to  twice  the 
depth. 
2.  The  cylinder  and  cone  have  volumes  in  the  ratio  3  :  2  and  lateral 

surfaces  in  the  ratio  2:3. 
4.  The  center  of  gravity  of  the  triangle  ABC. 


INDEX 


The  numbers  refer  to  the  pages. 


Acceleration,  along  a  straight  line, 
33. 
angular,  34. 

in  a  curved  path,  90,  91. 
Angle,  between  directed  lines  in 
space,  79,  80. 
between  two  plane  curves,  64. 
Approximate  value,  of  the  incre- 
ment of  a  function,   14,   15, 
118-120. 
Arc,  differential  of,  72. 

Continuous  function,  10,  113. 
Convergence  of  infinite  series,  107- 

111. 
Curvature,  73. 

center  and  circle  of,  75. 

direction  of,  67. 

radius  of,  74. 
Curve,  length  of,  70. 

slope  of,  11. 

Dependent  variables,  2,  115. 
Derivative,  12. 

directional,  129. 

higher,  28,  29,  114. 

of  a  function  of  several  vari- 
ables, 124-127. 

partial,  114. 
Differential,  of  arc,  72. 

of  a  constant,  20. 

of  a  fraction,  22. 

of  an  nth  power,  22. 

of  a  product,  21. 

of  a  sum,  20. 

total,  120,  121. 
Differentials,  15. 

exact,  130,  131. 

of  algebraic  functions,  19-31. 

of  transcendental  functions,  49- 
62. 

partial,  120. 
Differentiation,  of  algebraic  func- 
tions, 19-31 


Differentiation,  of  transcendental 

functions,  49-62. 
partial,  113-139. 
Directional  derivative,  129. 
Direction  cosines,  80,  81. 
Direction  of  curvature,  67. 
Divergence  of  infinite  series,  107- 

111. 

Exact  differentials,  130,  131. 
Exponential  functions,  56-62. 

Function,  continuous,  10,  113. 

discontinuous,  10. 

explicit,  1. 

implicit,  2,  127,  128. 

irrational,  2. 

of  one  variable,  1. 

of  several  variables,  113. 

rational,  2. 
Functions,  algebraic,  2,  19-31. 

exponential,  56-62. 

inverse  trigonometric,  54-56. 

logarithmic,  56-62. 

transcendental,  2,  49-62. 

trigonometric,  49-53. 
Functional  notation,  3. 

Geometrical  applications,  63-84. 

Implicit  functions,  2,  127. 
Increment,  10. 

of  a  function,  14,  15,  118,  119. 
Independent  variable,  2. 
Indeterminate  forms,  95-100. 
Infinitesimal,  7. 
Infinite  series,  106-112. 

convergence  and  divergence  of, 
107-111. 

Maclaurin's,  106. 

Taylor's,  106. 
Inflection,  67. 


161 


162 


INDEX 


Length  of  a  curve,  70. 
Limit,  of  a  function,  5. 

c  sin  9    An 
of  -j-,  49. 

Limits,  4-9. 

properties  of,  5,  6. 
Logarithms,  56,  58. 

natural,  58. 

Maclaurin's  series,  106. 
Maxima  and  minima,  exceptional 
types,  45,  46. 

method  of  finding,  42,  43. 

one  variable,  39-48. 

several  variables,  136-138. 
Mean  value  theorem,  101. 

Natural  logarithm,  58= 
Normal,  to  a  plane  curve,  63. 
to  a  surface,  133,  134. 

Partial  derivative,  114. 

geometrical    representation    of, 
116,  117. 
Partial,  differentiation,  113-139. 

differential,  120. 
Plane,  tangent,  134. 
Point  of  inflection,  67,  68. 
Polar  coordinates,  77-79. 
Power  series,  110,  111. 

operations  with,  111. 


Rate  of  change,  32. 
Rates,  32-38. 
related,  35. 
Related  rates,  35. 
Rolle's  theorem,  94. 

Series,  106-112. 

convergence  and  divergence  of, 
107-111. 

Maclaurin's,  106. 

power,  110,  111. 

Taylor's,  106. 
Sine  of  a  small  angle,  49. 
Slope  of  a  curve,  11. 
Speed,  85. 

Tangent  plane,  134. 
Tangent,  to  a  plane  curve,  63. 

to  a  space  curve,  81-83. 
Taylor's,  theorem,  102. 

series   106 
Total  differential,  120,  121. 

Variables,  change  of,  30,  127. 

dependent,  2,  115. 

independent,  2. 
Vector,  85. 

notation,  88. 
Velocities,  composition  of,  89,  90. 
Velocity,  components  of,  86,  87. 

along  a  curve,  85-89. 

along  a  straight  line,  32. 

angular,  34. 


INTEGRAL  CALCULUS 


BY 


H.  B.  PHILLIPS,  Ph.D. 

Assistant  Professor  of  Mathematics  in  the  Massachusetts 
Institute  of  Technology 


TOTAL  ISSUE  NINE  THOUSAND 


NEW   YORK 
JOHN    WILEY    &    SONS,  Inc. 

London:    CHAPMAN    &    HALL,    Limited 


Copyright,  1917, 

BY 

H.  B.  PHILLIPS 


Stanbope  jprcss 

F.    H.GILSON   COMPANY 

BOSTON,  U.S.A.  9_2] 


PREFACE 


This  text  on  Integral  Calculus  completes  the  course  in 
mathematics  begun  in  the  Analytic  Geometry  and  continued 
in  the  Differential  Calculus.  Throughout  this  course  I  have 
endeavored  to  encourage  individual  work  and  to  this  end 
have  presented  the  detailed  methods  and  formulas  rather 
as  suggestions  than  as  rules  necessarily  to  be  followed. 

The  book  contains  more  exercises  than  are  ordinarily 
needed.  As  material  for  review,  however,  a  supplementary 
list  of  exercises  is  placed  at  the  end  of  the  text. 

The  .appendix  contains  a  short  table  of  integrals  which 
includes  most  of  the  forms  occurring  in  the  exercises.  Through 
the  courtesy  of  Prof.  R.  G.  Hudson  I  have  taken  a  two-page 
table  of  natural  logarithms  from  his  Engineers'  Manual. 

I  am  indebted  to  Professors  H.  W.  Tyler,  C.  L.  E.  Moore, 
and  Joseph  Lipka  for  suggestions  and  assistance  in  preparing 
the  manuscript. 

H.  B.  PHILLIPS. 

Cambridge,   Mass. 
June,  1917. 


111 


CONTENTS 


Chapter  Pages 

I.   Integration 1-13 

II.   Formulas  and  Methods  of  Integration 14-34 

III.  Definite  Integrals 35-46 

IV.  Simple  Areas  and  Volumes 47-59 

V.   Other  Geometrical  Applications 60-69 

VI.   Mechanical  and  Physical  Applications 70-89 

VII.   Approximate  Methods 90-96 

VIII.   Double  Integration 97-111 

IX.   Triple  Integration 112-125 

X.   Differential  Equations 126-156 

Supplementary  Exercises 157-170 

Answers 171-185 

Table  of  Integrals 186-189 

Table  of  Natural  Logarithms 190-191 

Index 193-194 


INTEGRAL  CALCULUS 


CHAPTER  I 

INTEGRATION 

1.   Integral.  —  A    function  F  (x)    whose    differential    is 
equal  to  /  (x)  dx  is  called  an  integral  of  /  (x)  dx.     Such  a 

function  is  represented  by  the  notation    /  /  (x)  dx.     Thus 
F  (x)  =    if  (x)  dx,        dF  (x)  =  f  (x)  dx, 

are   by   definition   equivalent   equations.     The   process   of 
finding  an  integral  of  a  given  differential  is  called  integration. 
For  example,  since  d  (x2)  =  2  x  dx, 


s- 


2  x  dx  =  x2. 
Similarly, 

/  cos  x  dx  =  sin  x,  J  ex  dx  =  e*. 

The  test  of  integration  is  to  differentiate  the  integral.     If 
it  is  correct,  its  differential  must  be  the  expression  integrated. 
2.   Constant  of  Integration.  —  If  C  is  any  constant, 

d[F(x)  +  C]  =  dF(x). 

If  then  F  (x)  is  one  integral  of  a  given  differential,  F  (x)  +  C 
is  another.     For  example, 

/  2  x  dx  =  x2  +  C,  I  cos  x  dx  =  sin  x  +  C, 

where  C  is  any  constant. 


Integration 


Chap.  1 


We  shall  now  prove  that,  if  two  continuous  Junctions  of  one 
variable  have  the  same  differential,  their  difference  is  constant. 

Suppose  Fi  (x)  and  F2  (x)  are 
functions  having  the  same  differ- 
ential.    Then 

dFj(x)  =  dF2(x). 

Let  y  =  F2  (x)  —  Fx  (x)  and  plot 
the  locus  representing  y  as  a 
function  of  x.  The  slope  of  this 
locus  is 


o 


T 

e 

1 


X 


Fig.  2. 


dy  -_    dF2  (x)  -dFi  (x) 
dx  dx 


=  0. 


Since  the  slope  is  everywhere  zero,  the  locus  is  a  horizontal 
line.     The  equation  of  such  a  line  is  y  =  C.     Therefore, 

F2  (x)  -  F1  (x)  =  C, 

which  was  to  be  proved. 

If  then  F  (x)  is  one  continuous  integral  of  /  (x)  dx,  any 
other  continuous  integral  has  the  form 


/ 


/  (x)  dx  =  F  (x)  +  C. 


(2) 


Any  value  can  be  assigned  to  C.     It  is  called  an  arbitrary 
constant. 

3.   Formulas.  —  Let   a   and   n   be    constants,    u,    v,    w, 
variables. 

I.      /  du  ±  dv  ±  dw  =    I  du  db    I  dv  ±    l  dw. 

II.      /  a  du  =  a    I  du. 

/un+1 
un  du  = -— -  +  C,  if  n  is  not  —  1. 
n  +  1 

/  u_1  du  =  I  - —  =  In  u  +  C. 
J  J    u 


III 
IV 


Art.  3  Formulas  3 

These  formulas  are  proved  by  showing  that  the  differential 
of  the  right  member  is  equal  to  the  expression  under  the 
integral  sign.  Thus  to  prove  III  we  differentiate  the  right 
side  and  so  obtain 

7  /  "n+1    i   ^\      (n  +  1)  un  du  , 

+  C )  =  - — — f- =  un  du. 


\n+  1         /  n+  1 

Formula  I  expresses  that  the  integral  of  an  algebraic  sum 
of  differentials  is  obtained  by  integrating  them  separately 
and  adding  the  results. 

Formula  II  expresses  that  a  constant  factor  can  be  trans- 
ferred from  one  side  of  the  symbol  /  to  the  other  without 

changing  the  result.     A  variable  cannot  be  transferred  in  this 
way.     Thus  it  is  not  correct  to  write 

I  xdx  =  x    I  dx  =  x2. 


Example  1 .       I  x5  dx. 


Apply  Formula  III,  letting  u  =  x  and  n  =  5.     Then  dx  = 
du  and 


/ 


/>«5-|-l  /v.6 


,5(fa  =  _  +  C  =  -+C. 

Ex.  2.       LsVxdx. 
By  Formula  II  we  have 

ft  Vx  dx  =  3   Ac*  dx  =  ?y-+  C  -  2x*  +  C. 

Ex.  3.      f(x  -  1)  (x  +  2)  dx. 

We  expand  and  integrate  term  by  term. 

f(x  -l)(x  +  2)  dx  =   f(x2  +x-2)dx 

==  "3  X     "|     2  X"  —  .'        *  O  • 


Integration 


Chap.  1 


Ex.  4.      r 


x2  -  2  x  +  1 


x6 


dx. 


Dividing  by  x3  and  using  negative  exponents,  we  get 


/ 


x2  -2x  + 


x6 


-dx=  fix-1  -2x~2  +  x~3) 


dx 


=  ln  X  +  2X-1  -hx~2  +  C 


=  In  x  +  - 


x      2x' 


+  C. 


Ex.  5.      fV2 


x  +  1  dx. 


■  If  u  =  2  x  -\-  1,  du  =  2  dx.  We  therefore  place  a  factor 
2  before  dx  and  \  outside  the  integral  sign  to  compensate 
for  it. 


fV2 x  +  1  dx  =  h  f(2x  +  1)2  2dx  =  \   Cu^ 


du 


=  lj  +  C=|(2x  +  l)i  +  C. 


a;  da; 

x2  +  l 


Ex.  6.     J 

Apply  IV  with  u  =  x2  +  1.     Then  du  =  2  x  dx  and 

/x  dx       1    C2  x  dx      1    Cdu      1 ,         .    ~     ,       /  o  ,  -r  ,  ^ 

„     „        f  4  x  +  2  , 

Ex.  7.      /  s -—  dx. 

J  2x  —  1 


By  division,  we  find 

4x  +  2 
2x-  1 
Therefore 


=  2  + 


2x-  1 


/ISt^^/^+^t)^^2^21^^-1^0- 


Art.  4  Motion  of  a  Particle 

EXERCISES 
Find  the  values  of  the  following  integrals: 

1.  JV-3x.  +  5*)d*.  16-/(^fei 

2.  f(v-—^\dx.  17.     Cx  Va?  —  x2 dx. 


'•/(^+^K  is-  Si 


x-  dx 


+  x2 


4.  j(V2x-  -^=\dx.  19-  fx*Vx*-ldx. 

5.  JV5.(*  +  2«  +  l)dB.  20'  J^+oaj  +  d^' 

//    /-          /-\,  \    01  r    {2x-\-a)dx 

(V^-Vx)3dx.  21.  1                      — - 
^  V  x2  +  ax  +  6 

7.     f  s  (x  +  a)  (x  +  b)  dx.  22  f    j dt 

J  J  1  —  at" 

Q     r2x  +  3  ,  - 

8-   J  x~~dx'  23.  jKa2-^2)§^. 

f(x2  +  l)'(x2-2)  ^  ,  ,       v      xdx 

J § dx'  25.  f(2+ )     xdx    ■ 

J  *  J  \    ^2x2  +  l/2x2  +  l 


3'   JV2I  +  T'  28.   /(V2^_V2^r^. 
A      C  xdx  -r3_  o 

*-J*  +  2  29.   S*  +  2*dt- 

-        r       xdx  r 

J  v^zrf  30-  J  (x3_  i)2*^- 


4.  Motion  of  a  Particle.  —  Let  the  acceleration  of  a 
particle  moving  along  a  straight  line  be  a,  the  velocity  v, 
and  the  distance  passed  over  s.     Then, 

dv  ds 

a  = 

Consequently, 


a  =  m'      v  =  jt 


dv  =  adt,         ds  =  v  dt. 


6  Integration  Chap.  1 

If  then  a  is  a  known  function  of  the  time  or  a  constant, 

v  =  I adt  +  d,  s  =  fvdt  +  C2.  (4) 

If  the  particle  moves  along  a  curve  and  the  components  of 
velocity  or  acceleration  are  known,  each  coordinate  can  be 
found  in  a  similar  way. 

Example  1.  A  body  falls  from  rest  under  the  constant 
acceleration  of  gravity  g.  Find  its  velocity  and  the  distance 
traversed  as  functions  of  the  time  t. 

In  this  case 

dv 

a  = 

Hence 


a  =  Jt  = g- 


v  =   /  g  dt  =  gt  +  C. 


Since  the  body  starts  from  rest,  v  =  0  when  t  =  0.     These 

values  of_;w  and  t  must   satisfy  the  equation  v  =  gt  +  C. 

Hence 

0  =  g-0  +  C, 

ds 
whence    C  =  0  and  v  =  gt.     Since  v  =  —,    ds  =  gt  dt   and 


s  =   Jgtdt  +  C  =  i  gt2  +  C. 


When  t  =  0,  s  =  0.     Consequently,  C  =  0  and  s  =  \  gt2. 

Ex.  2.  A  projectile  is  fired  with  a  velocity  v0  in  a  direction 
making  an  angle  a  with  the  horizontal  plane.  Neglecting 
the  resistance  of  the  air,  find  its  motion. 

Pass  a'  vertical  plane  through  the  line  along  which  the 
particle  starts.  In  this  plane  take  the  starting  point  as 
origin,  the  horizontal  line  as  £-axis,  and  the  vertical  line  as 
?/-axis.  The  only  acceleration  is  that  of  gravity  acting 
downward  and  equal  to  g.     Hence 

—  =0  dLl=  -a  \ 

dt2         '  dt2  y' 


Art.  5 


Curves  with  a  Given  Slope 


1 

Integration  gives, 

When  <  =  0,  -t:  and  -77  are  the  components  of  v0.     Hence 
at  at 

Ci  =  v0  COSa,  C2  =  v0  sin  a, 

and 

c/x 
dl 
dy 
dt 


=  Vq  COS  a, 

=  Vq  sin  a  —  gt. 


Fig.  4. 


Integrating    again,    wc 
get 

X  =  v0t  cos  a, 

y  =  v0t  sin  a  —  §  jtf8; 

the  constants  being  zero  because  x  and  2/  are  zero  when 
(  =  0. 

5.    Curves  with  a  Given  Slope.  —  If  the  slope  of  a  curve 
is  a  given  function  of  x, 


dy 


-/to, 


then 
and 


dx 
<ty  =  /  to  dx 


2/ 


-//« 


dx  +  C 


is  the  equation  of  the  curve. 
Since  the  constant  can 
have  any  value,  there  are  an 
infinite  number  of  curves 
having  the  given  slope.  If 
the  curve  is  required  to  pass 
through  a  given  point  P,  the 

value  of  C  can  be  found  by  substituting  the  coordinates  of  P 

in  the  equation  after  integration. 


Fig.  5. 


8  Integration  Chap,  j 

Example  1.     Find  the  curve  passing  through  (1,  2)  witi 
slope  equal  to  2  x. 
In  this  case 


16. 
las 

on 


Hence 


^  =  2x.  I* 

dx  ^e 

to 

ai 


2/  =  I  2xdx  =  x2  -{-  C. 


Since  the  curve  passes  through   (1,  2),  the  values  x  —  1, 
y  =  2  must  satisfy  the  equation,  that  is 

2  =  1  +  0.  ' 

Consequently,  (7  =  1  and  2/  =  x2  +  1  is  the  equation  of  the 
curve. 

Ex.  2.     On  a  certain  curve 

d2y 


dx" 


=  x. 


=f&dx=fxdx  =  lx*+c- 


If  the  curve  passes  through  (—  2,  1)  and  has  at  that  point 
the  slope  —  2,  find  its  equation. 
By  integration  we  get 

dy 
dx 

At  (-  2,  1),  x  =  -  2  and  ^  =  -  2.     Hence 

dx 

-  2  =  2  +  C, 
or  C  =  —  4.     Consequently, 

2/  =  Ai  z2  -  4)  dx  =  J  x3  -  4  z  +  C. 

Since  the  curve  passes  through  (—  2,  1), 

1  =  - 1  +  s  +  a 

Consequently,  C  =  —  5§,  and 

2/  =  iz3-4x-5f 

is  the  equation  of  the  curve. 


Ill 


~t.  6 


Separation  of  the  Variables 


9 


- 


6.  Separation  of  the  Variables.  — ■  The  integration  formu- 
as  contain  only  one  variable.  If  a  differential  contains  two 
)r  more  variables,  it  must  be  reduced  to  a  form  in  which 
3ach  term  contains  a  single  variable.  If  this  cannot  be  done, 
ive  cannot  integrate  the  differential  by  our  present  methods. 

Example  1.  Find  the  curves  such  that  the  part  of  the 
tangent  included  between  the  coordinate  axes  is  bisected 
at  the  point  of  tangency. 

Let  P  (x,  y)  be  the  point  at  which  A B  is  tangent  to  the 
curve.     Since  P  is  the  middle 
point  of  AB, 

OA  =  2  y,         OB  =  2  x. 

The  slope  of  the  curve  at  P  is 

dy  OA  y 

dx~  ~  OB~  ~x" 

This  can  be  written 
•        dy     dx  =  Q 

y      x 

Since  each  term  contains  a  single  variable,  we  can  integrate 
ind  so  get 

In  y  -f-  In  x  =  C. 

This  is  equivalent  to 

In  xy  =  C. 
Hence 

xy  =  ec  =  k. 

C,  and  consequently  k,  can  have  any  value.  The  curves 
are  rectangular  hyperbolas  with  the  coordinate  axes  as 
asymptotes. 

Ex.  2.     According  to  Newton's  law  of  cooling, 

—  =  -  k  (0  -  a), 
at 

when  k  is  constant,  a  the  temperature  of  the  air,  and  6  the 
temperature  at  the  time  t  of  a  body  cooling  in  the  air.  Find 
6  as  a  function  of  t. 


10 


Integration 


Chap.  1 


Multiplying    by    dt    and    dividing    by   0  —  a,    Newton's 
equation  becomes 

=  —kdt. 

0  —  a 

Integrating  both  sides,  we  get 

In  (0  -  a)  =  -  kt  +  C. 
Hence 

$  -  a  =  e~kt+c  =  ece~kt. 

When  t  =  0,  let  0  =  0O.     Then 

do—  a  =  ece°  =  ec, 

6  -  a  =  (0O  -  a)  e~kt 


and  so 


is  the  equation  required. 

Ex.  3.  The  retarding  effect  of  fluid  friction  on  a  rotating 
disk  is  proportional  to  the  angular  speed  a>.  Find  w  as  a 
function  of  the  time  t. 

The  statement  means  that  the  rate  of  change  of  co  is  pro- 
portional to  co,  that  is, 

where  k  is  constant.     Separating  the  variables,  we  get 

—  =  kdt, 
co 


whence 
and 


In  co  =  kt  +  C, 


CO 


e^e 


Let  coo  be  the  value  of  co  when  t  =  0.     Then 

coo  =  ek'°ec  =  ec. 

Replacing  ec  by  co0,  the  previous  equation  becomes 

co  =  cooe*', 


which  is  the  result  required. 


Art.  6  -Separation  of  the  Variables  11 

Ex.  4.  A  cylindrical  tank  full  of  water  has  a  leak  at  the 
bottom.  Assuming  that  the  water  escapes  at  a  rate  pro- 
portional to  the  depth  and  that  ^  of  it  escapes  the  first 
day,  how  long  will  it  take  to  half  empty? 

Let  the  radius  of  the  tank  be  a,  its  height  h  and  the  depth 
of  the  water  after  t  days  x.  The  volume  of  the  water  at  any 
time  is  ira2x  and  its  rate  of  change 

9  dx 

This  is  assumed  to  be  proportional  to  x,  that  is, 

„  dx     , 
ira2  -77  =kx, 
dt 

where  k  is  constant.     Separating  the  variables, 

ira2  dx 


x 


=  kdt. 


Integration  gives 


wa2  In  x  =  kt  +  C. 

When  t  =  0  the  tank  is  full  and  x  =  h.     Hence 

ira2  In  /i  =  C. 

Subtracting  this  from  the  preceding  equation,  we  get 

ira2  In  t  =  kt. 
h 

When  t  =  1,  x  =  j%  h.     Consequently, 

ira2  In  x9^  =  k. 
When  x  —  \  h, 

ira2  In  -r       ,    x 

t  =  — j =  ; — \  =  6.57  days. 

k  In  T%  J 


12  Integration  *        Chap.  1 

EXERCISES 

1.  If  the  velocity  of  a  body  moving  along  a  line  is  v  =  2  t  +  3  t2, 
find  the  distance  traversed  between  t  =  2  and  t  =  5. 

2.  Find  the  distance  a  body  started  vertically  downward  with  a 
velocity  of  30  ft. /sec.  will  fall  in  the  time  t. 

3.  From  a  point  60  ft.  above  the  street  a  ball  is  thrown  vertically 
upward  with  a  speed  of  100  ft. /sec.  Find  its  height  as  a  function  of 
the  time.     Also  find  the  highest  point  reached. 

4.  A  rifle  ball  is  fired  through  a  3-inch  plank  the  resistance  of  which 
causes  a  negative  constant  acceleration.  If  its  velocity  on  entering 
the  plank  is  1000  ft. /sec.  and  on  leaving  it  500  ft. /sec,  how  long  does 
it  take  the  ball  to  pass  through? 

6.  A  particle  starts  at  (1,  2).  After  t  seconds  the  component  of  its 
velocity  parallel  to  the  x-axis  is  2  t  —  1  and  that  parallel  to  the  y-axis 
is  1  —  t.  Find  its  coordinates  as  functions  of  the  time.  Also  find  the 
equation  of  its  path. 

l/TJ  A  bullet  is  fired  at  a  velocity  of  3000  ft. /sec.  at  an  angle  of  45°  from 
a  point  100  ft.  above  the  ground.  Neglecting  the  resistance  of  the  air, 
find  where  the  bullet  will  strike  the  ground. 

7.  Find  the  motion  of  a  particle  started  from  the  origin  with  velocity 
Vo  in  the  vertical  direction,  if  its  acceleration  is  a  constant  K  in  a  direc- 
tion making  30°  with  the  horizontal  plane. 

8.  Find  the  equation  of  the  curve  with  slope  2  —  x  passing  through 
(1,  0). 

9.  Find  the  equation  of  the  curve  with  slope  equal  to  y  passing 
through  (0,  1). 

10.  On  a  certain  curve 

ax 

If  the  curve  passes  through  (1,  2),  find  its  lowest  point. 

11.  On  a  certain  curve 

*M  =  x-l. 

dx2 

If  the  curve  passes  through  (—  1,  1)  and  has  at  that  point  the  slope  2, 
find  its  equation. 

12.  On  a  certain  curve 

If  the  slope  is  —  1  at  x  =  0,  find  the  difference  of  the  ordinates  at  x  =  3 

and  x  =  4. 


Art.  6  Separation  of  the  Variables  13 

13.  The  pressure  of  the  air  p  and  altitude  above  sea  level  h  are  con- 
nected by  the  equation 

&  -  -  kv 

dh  kp' 

where  k  is  constant.     Show  that  p  =  poe-*\  when  p0is  the  pressure  at 
sea  level. 

14.  Radium  decomposes  at  a  rate  proportional  to  the  amount 
present.  If  half  the  original  quantity  disappears  in  1800  years,  what 
percentage  disappears  in  100  years?. 

15.  When  bacteria  grow  in  the  presence  of  unlimited  food,  they 
increase  at  a  rate  proportional  to  the  number  present.  Express  that 
number  as  a  function  of  the  time. 

16.  Cane  sugar  is  decomposed  into  other  substances  through  the 
presence  of  acids.  The  rate  at  which  the  process  takes  place  is  propor- 
tional to  the  mass  x  of  sugar  still  unchanged.  Show  that  x  =  ce-*'. 
What  does  c  represent? 

17.  The  rate  at  which  water  flows  from  a  small  opening  at  the 
bottom  of  a  tank  is  proportional  to  the  square  root  of  the  depth  of  the 
water.  If  half  the  water  flows  from  a  cylindrical  tank  in  5  minutes, 
find  the  time  required  to  empty  the  tank. 

18.  Solve  Ex.  17,  when  the  cylindrical  tank  is  replaced  by  a  conical 
funnel. 

19.  A  sum  of  money  is  placed  at  compound  interest  at  6  per  cent  per 
annum,  the  interest  being  added  to  the  principal  at  each  instant.  How 
many  years  will  be  required  for  the  sum  to  double? 

20.  The  amount  of  light  absorbed  in  penetrating  a  thin  sheet  of 
water  is  proportional  to  the  amount  falling  on  the  surface  and  approxi- 
mately proportional  to  the  thickness  of  the  sheet,  the  approximation 
increasing  as  the  thickness  approaches  zero.  Show  that  the  rate  of 
change  of  illumination  is  proportional  to  the  depth  and  so  find  the 
illumination  as  a  function  of  the  depth. 


CHAPTER  II 

FORMULAS  AND  METHODS  OF  INTEGRATION 

7.  Formulas.  —  The  following  is  a  short  list  of  integra- 
tion formulas.  In  these  u  is  any  variable  or  function  of  a 
single  variable  and  du  is  its  differential.  The  constant  is 
omitted  but  it  should  be  added  to  each  function  determined 
by  integration.  A  more  extended  list  of  formulas  is  given  in 
the  Appendix. 

/u1l+1 
undu  =  — r— ,  if  n  is  not  —  1. 
n  +  1 

II.     /  —  =  In  u. 
J    u 

III.  /  cos  udu  =  sin  u. 

IV.  /  sin  udu  =  —  cos  u. 
V.     /  sec2  u  du  =  tan  u. 

VI.  /  esc2  udu  =  —  cot  u. 

VII.  /  sec  u  tan  udu  =  sec  u. 

VIII.  I  esc  u  cot  u  du  =  —  esc  u. 

IX.  /  tan  udu  =  —  In  cos  u. 

X.  I  cot  u  du  =  In  sin  u. 

XI.     /  sec  u  du  =  In  (sec  u  +  tan  u). 

14 


Art.  8  Integration  by  Substitution  15 


XII.     /  esc  u  du  =  In  (esc  u  —  cot  it). 

•  f-A 

J  Va1 


(hi  .   _!« 

XIII.      /  — =:  =  sin  1  — 


,2  —  u~  a 


Xiv.     f    **=!tan-»H. 

J  u    -f-  a         a  a 


du  1        _x  u 


* 


XV.     f       du  Vr— 

XVI 


A** 


.      f   ,du         =  In  (u  +  Vu2  ±  a2). 
J  Vit2  ±  a2 

xvii.    r-*L-  =  im^.  +*±4m  ^^ 

J  u2  -  a2       2a       u  +  a  ^^  4^41 

XVIII.      fe"  <7u  =  e". 

Any  one  of  these  formulas  can  be  proved  by  showing  that 
the  differential  of  the  right  member  is  equal  to  the  expression 
under  the  integral  sign.     Thus  to  show  that 

/  sec  u  du  =  In  (sec  u  +  tan  u), 

we  note  that 

7 ,    z  .  N      (sec  u  tan  u  +  sec2  u)  du  7 

a  In  (sec  u  +  tan  u)  = r— =  sec  u  du. 

sec  u  +  tan  u    * 

8.  Integration  by  Substitution.  —  When  some  function 
of  the  variable  is  taken  as  u,  a  given  differential  may  assume 
the  form  of  the  differential  in  one  of  the  integration  formulas 
or  differ  from  such  form  only  by  a  constant  factor.  Inte- 
gration accomplished  in  this  way  is  called  integration  by 
substitution. 

Each  differential  is  the  product  of  a  function  of  u  by  du. 

More  errors  result  from  failing  to  pay  attention  to  the  du 

it 
*  In  Formulas  XIII  and  XV  it  is  assumed  that  sin-1  -   is  an  angle  in 

a 

it 
the  1st  or  4th  quadrant,  and  see-1  -  an  angle  in  the  1st  or  2nd  quadrant. 

In  other  cases  the  algebraic  sign  of  the  result  must  be  changed. 


16  Formulas  and  Methods  of  Integration  Chap.  2 

than  from  any  other  one  cause.  Thus  the  student  may 
carelessly  conclude  from  Formula  III  that  the  integral  of  a 
cosine  is  a  sine  and  so  write 


/ 


cos  2  x  dx  =  sin  2  x. 


If,  however,  we  let  2  x  =  u,  dx  is  not  du  but  \  du  and  so 
/  cos  2  x  dx  =  \    I  cos  u  du  =  \  sin  u  =  \  sin  2  x. 

Example  1.       /  sin3  x  cos  x  dx. 

If  we  let  u  =  sin  x,  du  =  cos  x  dx  and 
/  sin3  x  cos  x  dx  =    I  u3  du  =  \uA  +  C  =  \  sin4  x  +  C. 


„     rt        r  sin  ^  x  dx 
Ex.2.    Jj 


+  COS  ^  X 

We  observe  that  sin  J  x  dx  differs  only  by  a  constant 
fa  tor  from  the  differential  of  1  +  cos  |  x.     Hence  we  let 

u  =  1  +  cos  ^  x. 

Then  du  =  —  J  sin  |  x  dx,         sin  ^  x  dx  =  —  3  du, 

,  r  sinjxdx  _  Tdw  o1         .   n 

and  /  r-rJ — j—  =-3|  —  =-3lnw  +  C 

J  1  +  cos  §  x  j    w 

=  -  3  In  (1  +  cos  \  x)  +  C. 
Ex.  3.      /  (tan  x  +  sec  x)  sec  x  dx. 

Expanding  we  get 
I  (tan  x  +  sec  x)  sec  xdx  =  I  tan  x  sec  x  dx  +  /  sec2  x  dx 


Ex.  4.  r 


=  sec  x  +  tan  x  +  C. 
3dx 


V2-3x2 


Art.  8  Integration  by  Substitution  17 

This  resembles  the  integral  in  formula  XIII.    Let  u  =  x  V3, 

a  =  V'2.     Then  du  =  v^  dx  and 

f    3  — 

Zdx  /       '  V3  _  C        du 


r   sdx     =  I      V3     _   ,-  r 

J  V2  -  3  x2  "  J  Va2  -  u2  "     SJ  Va 


—  K? 


U        ~  rx    .      .  X  V3 


=  V3  sin"1-  +  C  =  V3  sin-1^-^  +  C. 
a  V2 


Ex  dt 


,  5.    / 


tV4:t2~9 


This  suggests  the  integral  in  formula  XV.     Let  u  =  2 1, 
a  =  3.     Then 

/eft  /»        2dt  r        du 

t  VA  t2  -  9      J  2tV±t2-9~~  J  u  Vu2  -  a2 


-  sec-1  -  +  C  =  o  sec-1  -=-  +  C 
a  a  6  6 


xdx 
Ex 


,  6.    /■ 


V2  x2  +  1 

This   may  suggest   formula   XVI.      If,  however,  we  let 

u  =  x  V2,  rfu  =  V2  dx,  which  is  not  a  constant  times  x  dx. 
We  should  let 

m  =  2x2  +  l. 

Then  xdx  =  \du  and 

/xdx      _  1   r<fa  _  1   r  _h 
V2x2  +  1  "  4 J  Vu~  ±JU 

=  |  V7*  +  C  =  |  V2  x2  +  1  +  C. 

Ex.  7.      /  etan  z  sec2  x  dx. 

If  u  =  tan  x,  by  formula  XVIII 

/  gtanx  sec2  xdx  =    f  6U  du  =  6U  +  C  =  eten  x  +  C. 


18  Formulas  and  Methods  of  Integration  Chap.  2 

EXERCISES 

Determine  the  values  of  the  following  integrals: 

1.     j  (sin  2  x  —  cos  3  x)  dx.  v"  21.     f  cos5  x  sin  x  dx. 


2 


sec2  x  dx 


'    /^PV5)*5'  22"    Jr+ 2 tans 

3.  C sin  (nt  + *)  dt.  23.     f  cos  2/^  . 
J  J  1  —  sin  2  x 

4.  I  sec2  £  0  d0.  24.     I  — — — '- — - 

J  J  1  +  tan  {ax) 

C     0    ,  e  ,„  nc     r      dx 

I  esc  -  cot  -  d0.  25.      I  — 

J         4         4  J  V3  — 


V3—  2  s2 
2dx 


/cos  0  sin  0  d0.  og  f  _~ 

^*  J  3  x2  +  4 

/  /*         dx 

cos2!;'  27-  J 


V3" 


x  v  dx 


/: 


/  7 

sin2  2  a;  28.     1         0         • 

J  12  t/2  +  3 
rcos  xaa;  , 

'    J"sm^*  29.     C        dX        • 

7  J  V7  x2  +  1 

/•sinxdx  ' 

J    cos5x  30.     j  ■ 

«       /•/       0  J\        0  ,„  ^  x  Va2x2-  y 

1.      I  (  esc  -  —  cot  -  1  csc  -  dd.  r      j 

V       2  2/  31.     I  -       .    2- 

/J  3  —  4  x2 
cos  (x2  —  1)  x  dr.  ^r 

J 


1     -  sin  3  x  7  J  V4  x1  —  3 

x  —  2)  d 

V4  —  x2 


r1  ■  —  dg. 

J      cos2  3  x  __       /•  (3  x  —  2)  dx 


4.     I  (sec  x—  l)2 


33. 


dx. 


/ 


t.     I dx.  J  v^  +  4 

./       cos  X  .    . 


2x  +  3 

/x2  + 
x  +  4 
x2—  { 
5x-  5 


dx. 


Tr1 — f^x. 
4x2—  5 

.  oc     r  5  x  —  2 

/*  (cos  x  +  sin  x)2  7  36.     I  =r 

7.  I r^ dx.  J  V3  x2  —  y 

J  sin  x 

/*     cos  x  dx 

8.  J  sin2  x  cos  x  dx.  '    J  V2  —  sin2  x 

//*sin  x  cos  x  dx 
tan3  x  sec2  x  dx.  "•    J      /^; 


/*  ,  «/v      /*  cos  x  dx 

20.     I  sec2  x  tan  x  dx.  39.     I  t 


sin2  x 
x 
-  sin2  x 


Art.  9        Integrals  Containing  a  Quadratic  Expression  19 


sec-xdx  Mm       f  ezx dx 

40. 


/soc-  x  dx  r  t 

tan  x  Vtaii-x—  1  '    Jl 

»*       /•sec  .r  tan  x  dx  .-       /vr 

41.      i  —  r-  49.      I  — 

J      Vs.M!2  X   -1-1  •/   C 


,3  x 


+  e 

•sec  .r  tan  .r  dx  rex  —  e~ 

Vsee2*  +  1  '    J  <?  +  e~ 

43.     1-71 n si"  «      r  e*da; 


cfc. 


+  c 


fc   Jx[4-  (Ins)8]!  51.  C- 

..       r    sin  x  cos  .r  r/.r 

J  V  cos2  x—  sin-  x  52.  f   e     (^x   . 

45.  t  — . 

46.  JV*2*dx.  '  J  Vl  -  ^2a/ 

47.  JV  +  «r«*)*<fc.  54.  fg^=? 


9.  Integrals  Containing  ax2  +  6z  +  c.  —  Integrals  con- 
taining a  quadratic  expression  ax2  +  bx  +  C  can  often  be 
reduced  to  manageable  form  by  completing  the  square  of 
ax2 .+  kc. 


Example  1.     /  r 


3  a;2  +  6  z  +  5 
Completing  the  square,  we  get 

3  x2  +  6  x  +  5  =  3  (x2  +  2  x  +  1)  +  2  =  3(x  +  l)2  +  2. 
If  then  u  =  (s  +  1)  V3, 

+  2 


r       rfg  r   d  (g  + 1)        _i_  r_j 

J  3x2  +  6x  +  5     J  3(x +  l)2  +  2  "  V3J  w2 


*  tan-'fr+^  +  C- 


£*.  2.  2  dx 


I 


V2  -  3  x  -  xl 


The  coefficient  of  x2  being  negative,  we  place  the  terms  x 
and  3  z  in  a  parenthesis  preceded  by  a  minus  sign.     Thus 

2  -  3  x  -  x2  =  2  -  (x2  +  3  x)  =  y-  -  (x  4  J )2. 


2 


20  Formulas  and  Methods  of  Integration  Chap.  2 

If  then,  u  =  x  +  f ,  we  have 

f         2dx  =2  f       du        =  28m-1  *+l  +  C 

J  V2-3x-x2        J  V\7-  -  u2  \  Vl7 

J  V4  a;2  +  4  x  +  2 

Since  the  numerator  contains  the  first  power  of  z,  we 
resolve  the  integral  into  two  parts, 

dx 


r  (2x-i)dx  =i  r  (8x+±)dx   _2  r 

J  V  4  x2  4-4x4-2     AJ  a/4 .r2 4-4 .t 4- 2        J  " 


V4x2+4z+2     4  J  V4z2+4z+2       «/  V4z2+4z+2 

In  the  first  integral  on  the  right  the  numerator  is  taken  equal 
to  the  differential  of  4  x2  +  4  x  +  2.  In  the  second  the 
numerator  is  dx.  The  outside  factors  J  and  —  2  are  chosen 
so  that  the  two  sides  of  the  equation  are  equal.  The  first 
integral  has  the  form 

i^  =  IV^  =  iV4:r2  +  4a;  +  2. 
Vu      z  * 

The  second  integral  is  evaluated  by  completing  the  square. 

The  final  result  is 


u 


i 


(2x—l)dx  1     /-. — 0    ,     . r-p. 

/                 =  =  -v4x2  +  4x  +  2 
V4x2  +  4z  +  2       2  

-ln(2z  +  l  +V4z2  +  4z4-2)  +C. 

EXERCISES 
1     C  dx  r  (2  x  +  5)  dx 

(2  re  —  1)  dx 


2.     f  ,         ^  8-   / 


3.     (  dx  -   --     .--.- 

J  V3.C2  4-4z  +  2  ,.      /•  (2x  +  3)dx. 


/•  xdx 

9-  J; 

da;.  ""'     *  (2z  +  l)V4z2+4z-l 


3x2--2z--2 
.0.     f 


J  (x2-  2x  +  3)f 
J  (z-3)V2^12x+15*  12'    J  V^zr^^ 


a;—  2 
(x  +  a)  (x  +  6)  ~ "   J    2  e2X  +  3  e*  -  1 


»/  (x  +  a)  (z  +  6)  J 


Art.  10  Integrals  of  Trigonometric  Functions  21 

10.  Integrals  of  Trigonometric  Functions.  —  A  power  of 
a  trigonometric  function  multiplied  by  its  differential  can  be 
integrated  by  Formula  I.     Thus,  if  u  =  tan  x, 

/  tan4  x  •  sec2  xdx  =    I  u*du  =  $  tan6  x  +  C. 

Differentials  can  often  be  reduced  to  the  above  form  by 
trigonometric  transformations.  This  is  illustrated  by  the 
following  examples. 

Example  1.       /  sin4  x  cos3  x  dx. 

If  we  take  cos  x  dx  as  du  and  use  the  relation  cos2  x  = 
1  —  sin2  x,  the  other  factors  can  be  expressed  in  terms  of 
sin  x  without  introducing  radicals.     Thus 

/  sin4 x  cos3 xdx  =  J  sin4 x  cos2 x  •  cos x dx 

=  I  sin4  x  (1  —  sin2  x)  d  sin  x  =  \  sin5  x  —  \  sin7  x  +  C. 

Ex.  2.       /  tan3  x  sec4  x  dx. 


s 


If  we  take  sec2  x  dx  as  du  and  use  the  relation  sec2  x  = 
1  +  tan2  x,  the  other  factors  can  be  expressed  in  terms  of 
u  =  tan  x  without  introducing  radicals.     Thus 

/  tan3 x  sec4 xdx  =  j  tan3 x  •  sec2 x  •  sec2 x dx 

=  J  tan3  x  (1  +  tan2  x)  d  tan  x 
=  \  tan4  x  +  i  tan6  x  +  C. 
Ex.  3.       /  tan3  x  sec3  x  dx. 


22  Formulas  and  Methods  of  Integration  Chap.  2 

If  we  take  tan  x  sec  x  dx  =  d  sec  x  as  du,  and  use  the  rela- 
tion tan2  x  =  sec2  x  —  1,  the  integral  takes  the  form 

/  tan3  x  sec3  xdx=  I  tan2  x  •  sec2  x  •  tan  a;  sec  x  dx 

=  I  (sec2  x  —  1)  sec2  a:  •  d  sec  # 

=  ^  sec5  x  —  J  sec3  £  +  C. 

Ex.  4.       I  sin  2  z  cos  3  x  dx. 

This  is  the  product  of  the  sine  of  one  angle  and  the  cosine 
of  another.  This  product  can  be  resolved  into  a  sum  or 
difference  by  the  formula 

sin  A  cos  B  =  J  [sin  (A  +  B)  +  sin  (A  -  B)]. 

Thus 

sin  2  #  cos  3  x  =  J  [sin  5  x  +  sin  ( —  x)  ] 

=  J  [sin  5  £  —  sin  #] 
Consequently, 

/  sin  2  x  cos  3  x  dx  =  \  I  (sin  5  x  —  sin  x)  da; 

=  —  TV  cos  5  x  +  £  cos  x  +  C. 
fe.  5.       /  tan5£c?:c. 

If  we  replace  tan2  x  by  sec2  x  —  1,  the  integral  becomes 

/  t&n5xdx  =  I  tan3  a;  (sec2  a;  —  1)  dx  =  J  tan4  a;  —  /  tan3#da\ 

The  integral   is  thus  made   to   depend   on  a   simpler  one 
/  tan3  x  dx.     Similarly, 

/  tan3  xdx  =  I  tan  x  (sec2  x  —  1)  dx  =  §  tan2  x  +  In  cos  x. 

* 

Hence  finally 

tan5 xdx  =  \  tan4 x  —  \  tan2 a;  —  In  cos x  +  C 


/< 


Art.  12 


Trigonometric  Substitutions 


23 


11.   Even  Powers  of  Sines  and  Cosines.  —  Integrals  of 
the  form 


/  sinm  x  cos71  x  dx, 


where  m  or  n  is  odd  can  be  evaluated  by  the  methods  of 
Art.  10.  If  both  m  and  n  are  even,  however,  those  methods 
fail.  In  that  case  we  can  evaluate  the  integral  by  the  use 
of  the  formulas 


sin2 

u 

— 

1  - 

cos  2 

u 

2 

cos2 

u 

= 

1  + 

COS  2 

u 

2 

r  r.ns 

a 

sin  2u 

2 


-  '\  r\v 


(ii) 


dx 


Example  1.      I  cosAxdx. 

By  the  above  formulas 

f     4    a         Cf      2   ^2^         f/l+cos2sY 

I  cos4  xdx  =  \  (cos2  x)2  02  =  1 1 5 

=  f  (i  +  §  cos  2  x  +  i  cos2  2  x) 
=  f  [i  +  icos2x  +  J(l  +  cos4x)]da; 
=  f  a;  +  J  sin  2  x  +  ^V  sin  4  x  +  C 
fe.  2.      /  cos2  x  sin2  a;  c?z. 

/  cos2 x sin2 xdx  =  I  \  sin2  2 x cfo  =  /  \  (1  —  cos 4 x)  dx 

=  J  05  —  5V  sin  4  x  +  C 

12.   Trigonometric    Substitutions.  —  Differentials    con- 
taining Va2  —  x2,   Va2  -f  z2,  or  Vz2  —  a2,  which  are  not 


24  Formulas  and  Methods  of  Integration  Chap.  2 

reduced  to  manageable  form  by  taking  the  radical  as  a  new 
variable,  can  often  be  integrated  by  one  of  the  following 
substitutions: 

For  Va2  -  x2,   let  x  =  a  sin  9.  '  ]  '■"''  ^  ' 

For  Va2  +  x2,  let  x  =  a  tan  0. 
For  Vx2  —  a2,  let  x  =  a  sec  0. 

Example  1.      /  Va2  —  x2  dx. 

Let  x  =  a  sin  0.     Then 

v  a2  —  a;2  =  a  cos  0,     dx  =  a  cos  0  d0. 
Consequently, 

/  Va2  —  a;2  cto  =  a2  I 

Since  x  =  a  sin  0, 

0  =  sin-1-> 
a 

Hence  finally 

fVa2  -  a; 

fe'2'     J\x2  +  a2y- 

If  we  let  a;  =  a  tan  0,  re2  +  a2  =  a2  sec2  6,  dx  =  a  sec2  0  c?0. 


z2cte  =  a2  I  cos26dd  =  ^^  +  isin2^  +  C. 


-  sin  2  0  =  sin  0  cos  0  = 5 

2  a2 


■2dz  =  ^sin-1-  + ~  Va2  -  x2  +  C. 


a 


<i# 


and 


r       dx  1    P  dd         1    T     9/1  Jn 

J  (^r^  =  -3J  ^rd  =  a*J  cos~ede 


2  a 


3  (0  +  sin  0  cos  0)  +  C. 


Since 


Hence 


x  ax 

x  =  a  tan  0,     0  =  tan-1  - ,     sin  0  cos  0  =  -=— ; — - 

a  a2  +  n- 

J  (x2  +  a2)2  =  ^L^^a  +  ^T^J  +  C* 


Art.  12  Trigonometric  Substitutions  25 

EXERCISES 

1.  f  sin3  xdx.     '  21.  J  sin2  ax  dx. 

2.  J  cos6 xdx.  22.  J  cos2  ax  dx. 

3.  f  (cos  x  4-  sin  x)3  dx.  23.  J  cos2  x  sin4  x  dx. 

4.  f  cos2  x  sin3  xdx.                            24.  J  cos4  \x  sin2 1  xdx. 
6.     (sin4  \  x  cos6  ^  x  dx.                     25.  J  sin6  x  dx. 

6.     fsin6  3  0  cos3  3  6  dd.  26.     ( j-^ 


dx 


sinx 
dx 


7.     f  (cos2 0  —  sin2 0)  sin 0 dd.  27«    J  r+~cosx ' 

L     fcos^xdx.  2g>    Vl+sin0d0. 
J  1  —  sin  x 

^     r cos2  xdx.  29>     fv^^dx. 
J       sin  x  J 

J     cos  0  »/ 


' 


x2  dx 


J  sec4  x  dx.  31. 

J  Vx2  +  a2 

2.     Ccsc"ydy.  32.     f — — 3- 

^  ^  (x2-a2)* 

.     j  tan2  x  dx.  gg.      f         "X  _ 

,  .    ,         ,  _  J  x  Va2  —  x2 
fsec30  +  tan30 

4'    J    sec  0  +  tan  0    ^  34.     C  d*  2 

C  x  ^2  ax  " '  x<! 

6.    J  tan  i  x  sec3  \  x  dx.  r      xdx 

35.     I  

/•s      ( rt%  —    o 
tan5  2  x  sec3  2  x  dx. 


(a2  —  x2) 


r  -  36.     fx3VxM:~a2dx 

7.     j    cot3 xdx. 

■/  /»  dr. 


C 


8.     f  tan7 xdx.  J  x2  Vx2  +  a2 

/cos2  x  dx  38.     f  Vx2—  4x  +  5 

sin6x 


dx. 
(x2  —  x)  dx 


20.    J  sec3  x  esc  x  dx.  39-    J  V2  —  2  x  —  4  x' 


26  Formulas  and  Methods  of  Integration  Chap.  2 

13.   Integration  of  Rational  Fractions.  —  A  fraction,  such 
as 

x3  +  3  x 


x2  -  2x-  3 


whose  numerator  and  denominator  are  polynomials  is  called 
a  rational  fraction. 

If  the  degree  of  the  numerator  is  equal  to  or  greater  than 
that  of  the  denominator,  the  fraction  should  be  reduced  by 
division.     Thus 

x3  +  9x  +  12  10x  +  6 

=  x  +  I  + 


x2-  2x-  3  '        '  x2-2x-f  3 

A  fraction  with  numerator  of  lower  degree  than  its  denomi- 
nator can  be  resolved  into  a  sum  of  partial  fractions  with 
denominators  that  are  factors  of  the  original  denominator. 
Thus 

10x  +  6  10x  +  6  9       .       1 

+ 


x2  -  2  x  -  3      (x  -  3)  (x  +  1)      x  -  3  '  x  +  1 

These  fractions  can  often  be  found  by  trial.  If  not,  pro- 
ceed as  in  the  following  examples. 

Case  1.  Factors  of  the  denominator  all  of  the  first  degree 
and  none  repeated. 

x4  +  2  x  +  6   , 
Ex.  1.       I     „  .     n — -=-  dx. 


•-1-  f. 


x3  +  x2  —  2  x 

Dividing  numerator  by  denominator,  we  get 

x4  +  2x  +  6  ,    ,        3x2  +  6 

=  x  —  1  + 


x3  +  x2  —  2  x  x3  +  x2  —  2  x 

3x2  +  6 


=  x-  1  + 


x  (x  -  1)  (x  +  2) 


Assume 


3x2  +  6  A   .      B      .      C 

—  ~  +  1 T  + 


x  (x  -  1)  (x  +  2)       x    '  x  -  1   '  x  +  2 
The  two  sides  of  this  equation  are  merely  different  ways  of 


Art.  13  Integration  of  Rational  Fractions  27 

writing  the  same  function.     If  then  we  clear  of  fractions,  the 
two  sides  of  the  resulting  equation 

3  x1  +  6  =  A  (x  -  1)  (x  +  2)  +  Bx  (x  +  2)  +  Cx  (x  - 1) 
=  {A  +  B  +  C)x2+(A  +  2B  -C)x-2A 

are  identical.     That  is 

A+B  +  C  =  3,     A+2B-C  =  0,     -2A  =  6. 

Solving  these  equations,  we  get 

A  =  -  3,     B  =  3,     C  =  3. 

Conversely,  if  A,  B,  C,  have  these  values,  the  above  equa- 
tions are  identically  satisfied.     Therefore 

3 


dx 


Jr>  +  z2-2x  J\x  x^x-l^x  +  2, 

=  \  x2  -  x  -  3  In  x  +  3  In  (x  -  1)  +  3  In  (x  +  2)  +  C 

1  2              Q1    (a?  -  1)  (a;  +  2)   ,   n 
=  jz  x2  —  x  +  3  In h  C 

2  a; 

The  constants  can  often  be  determined  more  easily  by 
substituting  particular  values  for  x  on  the  two  sides  of  the 
equation.     Thus,  the  equation  above, 

3  x2  +  6  =  A  (x  -  1)  (x  +  2)  +  Bx  (x  +  2)  +  Cx  (x  -  1) 

is  an  identity,  that  is,  it  is  satisfied  by  all  values  of  x.     In 
particular,  if  x  =  0,  it  becomes 

6=  -2  A, 

whence   A  =  —  3.     Similarly,   by    substituting  x  =  1   and 

x  =  —  2,  we  get 

9  =  35,     18  =  6  C, 

whence  5  =  3,         C  =  3. 

Case  2.  Factors  of  the  denominator  all  of  first  degree 
but  some  repeated. 

(8  x3  +  7)  dx 


Ex.  2 


■  f 


(x  +  l)(2;r  +  l)3 


28  Formulas  and  Methods  of  Integration  Chap.  2 

Assume 

8z3  +  7  A      .         B         ,         C  D 


(z  +  l)(2z  +  l)3      x+1      (2z+l)3  '   (2x+l)2  '  2x  +  V 

Corresponding  to  the  repeated  factor  (2  x  +  l)3,  we  thus 
introduce  fractions  with  (2  x  +  l)3  and  all  lower  powers  as 
denominators.     Clearing  and  solving  as  before,  we  find 

A  =  1,         B  =  12,         C  =  -  6,        D  =  0. 
Hence 

r     8x3+7        =  rr  i    ,     12 6    1 

J  (x+l)(2x+l)3aX     J  |x  +  l      (2s  +  l)8     Qx+l)*]™ 

Case  3.     Denominator  containing  factors  of  the  second 
degree  but  none  repeated. 


:.3.    f 


htX.  d.       I         ^      z       dx. 

x3  —  1 


The  factors  of  the  denominator  are  x  —  1  and  x2  +  x  +  1. 
Assume 

4z2  +  z+l         A  £z  +  C 


a?  -  1  x  -  1  '  a?2  +  x  +  1 

With  the  quadratic  denominator  z2  -+  #  +  1,  we  thus  use  a 
numerator  that  is  not  a  single  constant  but  a  linear  function 
Bx  +  C.     Clearing  fractions  and  solving  for  A,  B,C,  we  find 

A  =  2,        £  =  2,        C  =  1. 

Therefore 

J       a:3  —  1  J  \x  —  1      x2  +  x  +  1/ 

=  2  In  (x  -  1)  +  In  Or2  +  x  +  1)  +  C. 

Case  4.     Denominator  containing  factors  of  the  second 
degree,  some  being  repeated. 

Ex.  4.      /  — ,     ',   ,NO dx. 


x  (x2  +  l)2 


Art.  14  Integrals  29 

Assume 

x3  +  1         A   .    Bx  +  C    ,Di  +  l! 

=  ™  "T  /    o    ,     no  + 


x  (x2  +  l)2       a:    '    (a:2  +  l)2  '    x2  +  1 

Corresponding  to  the  repeated  second  degree  factor  (x2  +  l)2, 
we  introduce  partial  fractions  having  as  denominators 
(xa  +  l)2  and  all  lower  powers  of  x2  +  1,  the  numerators 
being  all  of  first  degree.  Clearing  fractions  and  solving  for 
A,  B,C,  D,  E,  we  find 

A  =  l,     B=-l,     C=-l,     D=-l,    E=l. 

Hence 

r  a^  + 1      =  f  f1     x  +  i      x-i , 

J  x  (x2  +  l)2  J  U      02  +  I)2      x2  +  1 

=  In     .  -f  x  tan-1  x  —  0  .  2        .  +  C. 

Vz2  +1      2  2  (z2  +  1) 

p 

14.   Integrals  Containing  (ax  +  b)'1.  —  Integrals  contain- 

p 
ing  (ax  +  6)9  can  be  rationalized  by  the  substitution 

ax  -\-b  =  zq. 

If  several  fractional  powers  of  the  same  linear  function 
ax  +  b  occur,  the  substitution 

ax  +  b  =  zn 

may  be  used,  n  being  so  chosen  that  all  the  roots  can  be 
extracted. 

Example  1.      /  =• 

J  1  +  Vi 

Let  z  =  z2.     Then  dr  =  2  2  dz  and 

f_dx__  =  r»2zrfg  _   /7 2 

J  l  +  Vx~J  1  +  *     J  V        1 


dz 


Vi      J   1  +  z      J  \         1  +  z> 
=  2z-  21n(l  +z)  +C 
=  2Vx-21n(l  +  Vs)+C. 


fc  f(2x-3M 

J  (2x-  3^  4- : 


(2a;  -3)3  +  1 


30 


Formulas  and  Methods  of  Integration 


Chap.  2    lit 

To    rationalize    both    (2  x  -  3)*    and    (2  x  -  3)3,    let 
2x-3  =  z6.    Then 


(2  a; -3)*  da; 


dz 


'z7      z5  .  z3 


=  3(-y  -  -  +-  -  z  +  tan"1  z)+C 

=  ?  (2a;  -  3)*-  f  (2a;  -  3)*  +  (2a;  -  3)* 
-  3  (2  a;  -  3)  *  +  tan"1  (2  x  -  3)*  +  & 


EXERCISES 


2-/ 


x3  +  #2 


3x  +  2 


dx. 


dx. 


2a;  +  3 

x2  +  x 

x2  +  1 

x  (x2  —  1) 

.    I  -. — 5 dx. 

./  4x3—  x 


4 


dx. 


x  dx 


8./ 


(x  +  1)  (x  +  3)  (x  +  5) 

16xdx 
(2x-l)(2x-3)(2x-5) 


x3  +  1 


dx. 


X" 

x2  dx 


(x  +  1)  (x-  l)2 


f       dx 

J  (x2-  l)5 


*  jew* 

13.  f4ia. 

J  (x2—  4)2 


x4dx 


*  J  x3- 

*  J  x^= 


X4 


r3  +  l 
x2  dx 


dx 


9 


X3  +  x2  —  1 

2  x2  +  x  —  2 
(x2-l)2     dX- 

x4  +  24  x2  -  8  x 


rix1  +  x- 

'  J  '   (x2  -  1 

'  J        (x3--  8)2 

20.  f(£+ill^. 
J  x 

21.  p'7^  +  1 

22.  fxVax~+bdx 
Vx+2— 1 


dx. 


dx. 


23.  / 

24-/^ 
25.  / 


a:  +  3 
dx 


dx. 


1)   (^    +    1) 

dx 


Vz  +  1  -  ^x  —  1 


rt.  15  Integration  by  Parts  31 

15.   Integration  by  Parts.  —  From  the  formula 

d  (uv)  =  udv  -f  udv 
re  get 

udv  =  d  (uv)  —  v du, 
/hence 

J  udv  =  uv  —    j  v du.  (15) 

If  j  vdu  is  known  this  gives  j  v  du.     Integration  by  the 
lse  of  this  formula  is  called  integration  by  parts. 

Example  1.       /  In  x  dx. 

Let  u  =  In  x,  dv  =  dx.     Then  du  =  — ,  v  =  x,  and 

a; 

I  In  x  dx  =  In  x  •  x  —   /  x  •  — 
=  a;(lna;-  1)  +C. 
Ex.  2.       /  x2  sin  x  dx. 

Let  w  =  x2  and  cfa  =  sin  x  dx.     Then   du  =  2  x  dx,  v  =* 
—  cos  x,  and 

x2  sin  xdx  =  —  x2  cos  x  +  /  2  x  cos  x  dx. 


/  x2  sin  x  dx  =  —  x2  cos  x  +  I ' 


A  second  integration  by  parts  with  w  =  2  x,  dv  =  cos  x  dx 
gives 

I  2  x  cos  x  dx  =  2  x  sin  x  —  /  2  sin  x  dx 

=  2  x  sin  x  +  2  cos  x  -f-  C. 
Hence  finally 

/  x2  sin  xdx  =  —  x2  cos  x  +  2  x  sin  x  +  2  cos  x  +  C. 


32  Formulas  and  Methods  of  Integration  Chap.  2 

The  method  of  integration  by  parts  applies  particularly 
to  functions  that  are  simplified  by  differentiation,  like  In  x, 
or  to  products  of  functions  of  different  classes,  like  x  sin  x. 
In  applying  the  method  the  given  differential  must  be  re- 
solved into  a  product  u  •  dv.  The  part  called  dv  must  have 
a  known  integral  and  the  part  called  u  should  usually  be 
simplified  by  differentiation. 

Sometimes  after  integration  by  parts  a  multiple  of  the 
original  differential  appears  on  the  right  side  of  the  equation. 
It  can  be  transposed  to  the  other  side  and  the  integral  can 
be  solved  for  algebraically.  This  is  shown  in  the  following 
examples. 

Ex.  3.  Va2  -  x2  dx. 


:.  3.       fVaF^ 


Integrating  by  parts  with  u  =  Va2  —  x2,  dv  =  dx,  we  get 

—  x2  dx 


/Va2  —  x2 dx  =  x  Va2  —  x2  —    /      . 
J  Va2- 


x2 


Adding  a2  to  the  numerator  of  the  integral  and  subtracting 
an  equivalent  integral,  this  becomes 


/Va2  —  x2dx  =  x  Va2  —  x2—    I     ,  dx-\-a2  I     , 

J  Va2  -  x2  J  Va2  - 

=  x  Va2  —  x2  —  I  Va2  —  x2 dx  +  a2  /  —7= 

J  J  Va 

Transposing   /  ^«2  ~  &  dx  and  dividing  by  2,  we  get 


x2 


2-x2 


i 


Va2  -x2dx  =  %  Va2  -  x2  +  %■  sin"1  -  +  C. 

£  —  a 


Ex.  4.       /  eax  cos  bx  dx. 

Integrating  by  parts  with  u  —  e?x,  dv  =  cos  bx  dx,  we  get 

C           -l     7        e°x  sin  bx      a   C  M       ,     , 
I  eax  cos  bx  dx  =  r t   /  e    sin  bx  dx. 


Art.  16  Reduction  Formulas  33 

Integrating   by   parts  again   with    u  =  enx,   dv  —  sin  bx  dx, 
this  becomes 

/,     .       eax  sin  bx     a\     eax  cos  bx  .  a   /'  „    .    ,     , 
eax  cos  bx  dx  = 7 7 7 \--r  J  eax  sin  bx  dx 

(b  sin  bx  +  a  cos  6x\       a2    f       .    ,     7 
=  cax  ( 7^ 1  —  t  2   /  e    sin  ^  «^* 

62 
Transposing  the  last  integral  and  dividing  by  1  +  -^  this 

gives 

7     7  /6  sin  bx  -\-  a  cos  6z\ 

eax  cos  bx  dx  =  eax[  — 


/< 


a2  +  b2  J 

16.  Reduction  Formulas.  —  Integration  by  parts  is  often 
used  to  make  an  integral  depend  on  a  simpler  one  and  so  to 
obtain  a  formula  by  repeated  application  of  which  the  given 
integral  can  be  determined. 

To  illustrate  this  take  the  integral 


/ 


sinn  x  dx, 


wnere  n  is  a  positive  integer.     Integrating  by  parts  with 
a  =  sinn_1  x,  du  =  sin  x  dx,  we  get 

/  sin"  x  dx  =  —  sin"-1  x  cos  x+  /  (n  —  1)  sinn~2  x  cos2  x  dx 

=  —  sin"-1  x  cos  z  +  (n—  1)  l  sinn-2x(l  —  sin2x)  dx 
=  —  sin"-1  z  cos  £  +  (n—  1)  I  smn~2xdx 

—  (n—  1)  /  sin"  xdx. 

Transposing  the  last  integral  and  dividing  by  n,  we  get 

/•  «     7            sin"-1  x  cos  x  ,  n  —  1  C  .  _.     , 
smn  xdx  = 1 /  sin"--2 x dx. 
n                   n    J 

By   successive   application   of   this   formula   we   can  make 
/  sin"  xdx  depend  on  /  dx  or  /  sin  x  dx  according  as  n  is 
even  or  odd. 


34 


Formulas  and  Methods  of  Integration 


Chap.  2 


Example.      I  sin6  x  dx. 


By  the  formula  just  proved 


/ 


sin6  x  dx  =  — 


sm°  x  cos  x 


6 


fsh 


+  ^    f  sin4  x  dx 


sin5  x  cos  x      5 
~6  6 


sin3  x  cos  x      3    r  .  0        ' 
4 *~  i  /  sln  x  dx 


sin5  a:  cos  x       5    .  ,  5    .  „    5      ,   ~ 

—  —  sin*5  a;  cos  x  —  77;  sin  x  cos  z  +  -7  x  +  C - . 

0  J4  lb  lb 


vV 


1.  i  x  cos  2  x  dx. 

2.  J  In  x  •  x  dx. 

3.  J  sin-1  x  dx. 

4.  Is  tan-1  x  dx. 

6.  fin  (x  +  Va2  +  s2)  dx. 

_  /*  In  x  dx 

6.  .- 

J  Vz—  1 

7.  Jin  (In  x)^. 

8.  J  x2  sec-1  x  dx. 

9.  fe-xln(cx  +  l)dx. 

10.  f  x2  ex  dx. 

20.  Prove  the  formula 


EXERCISES 


11.  \x3e~xdx. 

12.  J" (a;—  l)2sin(2x)dx. 

13.  fVx2—  a?dx. 

14.  fVtf~+tfdx. 

15.  J  e2X  sin  3  x  dx. 

16.  I  ex  cos  x  dx. 

17.  j  e_z  sin  2  x  dx. 

18.  fsec30d0. 

19.  I  sin  2  x  cos  3  x  dx. 


Jn  .  N  ,         sec*1-2  x  tan  x      n  —  2  /•      __„  .  . 
secn  (x)  dx  = r   I  sec71-2  (x)  dx. 
n  —  1             n—  1  J 

and  use  it  to  integrate    J  sec5  x  dx. 

21.   Prove  the  formula 


f(a2~  x2)ndx 


x(a2  —  xn)n  .      2  na2 


2n  +  1       '    2n  - 
and  use  it  to  integrate  J  (a2  —  x2)^  dx. 


+  2TTT/(o2-a;2)"-,da! 


CHAPTER  III 


a   Xi  x2  x3  x4  x5  x6  Xj  b 


DEFINITE   INTEGRALS 

17.  Summation.  —  Between  x  =  a  and  x  =  6  let  /(x)  be 
a  continuous  function  of  x.  Divide  the  interval  between  a 
and  b  into  any  number  of  equal  parts  Ax  and  let  Xi,  x2,  .  .  . 
xn,  be  the  points  of  division.     Form  the  sum 

f(a)Ax+f(x1)Ax+f(x2)Ax  +  •  •  •  +f(xn)kx. 

This  sum  is  represented  by 
the  notation 

XI /MA* 

Since  /  (a),  /(x0,  /  (x2), 
etc.,  are  the  ordinates  of 
the  curve  y  =  f  (x)  at 
x  =  *i,  £2,  etc.,  the  terms 
/  (a)  Ax,  /  (x>)  Ax,  /  (x,)  Ax,  FlG-  17a- 

etc.,   represent    the   areas  of   the   rectangles  in   Fig.    17a, 

and  ]x  /  (x)  Ax  is  the  sum  of  those  rectangles. 

Example  1.     Find  the  value  of  ^ ~  x2  Ax  when  Ax  =  J. 

The  interval  between  1  and  2 
is  divided  into  parts  of  length 
Ax  =  J.  The  points  of  division 
are  If,  lj,  If.     Therefore 

2}'x2Ax  =  V-Ax+  (f)2Ax  + 
(f)2Ax  +  (l)2Ax 
=  *j-Ax  =  %3  -l  =  1.97. 

Ex.   2.     Find  approximately  the  area  bounded  by  the 
x-axis,  the  curve  y  =  Vx,  and  the  ordinates  x  =  2,  x  =  4. 
From  Fig.  176  it  appears  that  a  fairly  good  approxima- 

35 


Fig.  176. 


36 


Definite  Integrals 


Chap.  3 


tion  will  be  obtained  by  dividing  the  interval  between  2  and 
4  into  10  parts  each  of  length  0.2.  The  value  of  the  area 
thus  obtained  is 

^ViAz=(v/2+V2^+V2^+  •  •  •  +^3^8)  (0.2)  =3.39. 

The  area  correct  to  two  decimals  (given  by  the  method  of 
Art.  20)  is  3.45. 

18.   Definite  and  Indefinite  Integrals.  —  If  we  increase 
indefinitely  the  number  of  parts  into  which  b  —  a  is  divided, 

the   intervals   Ax   approach   zero  and    V  /  (x)  Ax  usually 

approaches  a  limit.  This  limit  is  called  the  definite  integral 
of  /  (x)  dx  between  x  =  a  and  x  =  b.     It  is  represented  by 

/    /  Or)  dx.     That  is 

/  (x)  dx  =  lim  Z.  f  (x)  Ax.  (18) 


the  notation 


£ 


Aj  =  0 


The  number  a  is  called  the  lower  limit,  b  the  upper  limit  of 
the  integral. 

In  contradistinction  to  the  definite  integral  (which  has  a 
definite  value),  the  integral  that  we  have  previously  used 
(which  contains  an  undetermined  constant)  is  called  an  in- 
definite integral.  The 
connection  between  the 
two  integrals  will  be 
shown  in  Art.  21. 

19.  Geometrical 
Representation. — If 
the  curve  y  =  f  (x)  lies 
above  the  x-axis  and 
a  <  b,  as  in  Fig.  17a, 

Jf  (x)  dx  represents 
a 

the  limit  approached  by  the  sum  of  the  inscribed  rectangles 
and  that  limit  is  the  area  between  x  =  a  and  x  =  b  bounded 
by  the  curve  and  the  x-axis. 


Fig.  19a. 


irt.  19. 


Geometrical  Representation 


37 


At  a  point  below  the  z-axis  the  ordinate  /  (x)  is  negative 
md  so  the  product  /  (x)  Az  is  the  negative  of  the  area  of  the 
:orresponding  rectangle.     Therefore  (Fig.  19a) 

2j  af  (x)  <±x  =  (sum  of  rectangles  above  OX) 

—  (sum  of  rectangles  below  OX), 
ind  in  the  limit 


f 


f  (x)  dx  =  (area  above  OX)  —  (area  below  OX)  (19a) 


Y 

r- 

0 

1 

> 

/               ax 

Fig.  Kb. 

If,  however,  a  >  b,  as  in  Fig.  196,  x  decreases  as  we  pass 
;rom  a  to  b,  \x  is  negative  and  instead  of  the  above  equation 
we  have 

/    f  (x)  dx  =  (area  below  OX)  —  (area  above  OX) .     (19b) 


Example  1.     Show  graphically  that 

The  curve  y  =  sin3  x  is 
hown  in  Fig.  19c.  Be- 
tween x  =  0  and  x  =  2ir  the 
areas  above  and  below  the 
x-axis  are  equal.     Hence 

Jr*27T 
sin3  x  dx  =  A  i  —  A2  =  0. 
o 

Ex.  2.     Show  that 


f 

Jo 


sin3  x  dx  =  0. 


u 


IT 


2tt 


X 


Fig.  19c. 


j    e~x2dx  =  2  Je-*2dx. 


38  Definite  Integrals  Chap.  5 

The  curve  y  =  e~z2  is  shown  in  Fig.  19a.     It  is  symmetrica 
Y  with   respect  to  the   y-axis 

The  area  between  x  =  —  ] 
and  x  =  0  is  therefore  equa 
to  that  between  x  =  0  anc 
x  =  1.     Consequently 


Fig.  19d. 


J '  e~*  dx  =  Ai  +  A2  =  2A2 


EXERCISES 

Find  the  values  of  the  following  sums: 
1.    ^    xAx,         Ax  =  £. 

™Ax 
—  > 
a; 


Ax  =  1. 


3.  2>       VxAx,         Ax  =  \. 

4.  Show  that 


X 


7T 


sin  x  Ax  =  1  —  cos  - 
o  6 


IT 


approximately.     Use  a  table  of  natural  sines  and  take  Ax  =  ^ 

5.    Calculate  ir  approximately  by  the  formula 

i     Ax 


TT 


■*Xt 


+  x2 


Ax  =  0.1. 


6.    Find  correct  to  one  decimal  the  area  bounded  by  the  parabola 
y  =  x2,  the  x-axis,  and  the  ordinates  x  =  0,  x  =  2.     The  exact  area 


is  f . 


7.    Find  correct  to  one  decimal  the  area  of  the  circle  x2  +  y2  =  4. 
By  representing  the  integrals  as  areas  prove  graphically  the  following 
equations : 


8.     C  sin  (2  x)  dx  =  0. 

Jo 

Jo 
10.      P 


cos7  x  dx  =  0. 


sin5  x  dx 


=  2f2si 

Jo 


sin5  x  dx. 


Art.  20 


Derivative  of  Area 


39 


+°     x  dx 


+  x4 


„  i 
*£tt*=2.Ct 


dx 


+  x* 
f  (x)  dx  =  j    /  (a  —  x)  dx. 

20.   Derivative  of  Area.  —  The  area  A  bounded  by  a 
curve 

y=f(x), 

a  fixed  ordinate  x  =  a,  and  a  movable  ordinate  MP,  is  a 
function  of  the  abscissa  x  of  the  movable  ordinate. 

Let  x  change  to  x  +  Ax.    r 
The  increment  of  area  is 

1  AA  =  MPQN. 

Construct  the  rectangle 
MP'Q'N  equal  in  area  to 
MPQN.  If  some  of  the 
points  of  the  arc  PQ  are 
above  P'Q',  others  must  be  ~o 
below  to  make  MPQN  and 
*  MP'Q'N  equal.  Hence  P'Q' 
intersects  PQ  at  some  point  R.  Let  yf  be  the  ordinate  of  R. 
Then  y'  is  the  altitude  of  MP'Q'N  and  so 


Fig.  20. 


A  A  =  AiTW  =  MP'Q'N  =  y'  Ax. 


Consequently 


AA 
Ax 


=  y 


When  Ax  approaches  zero,  if  the  curve  is  continuous,  y' 
approaches  y.     Therefore  in  the  limit 

^  =  y=f(x).  (20a) 

Let  the  indefinite  integral  of  /  (x)  dx  be 
I  ff(x)dx  =  F(x)+C. 


40  Definite  Integrals  Chap.  3 

From  equation  (20a)  we  then  have 

A  =  Cf(x)dx  =  F(x)  +  C. 

The  area  is  zero  when  x  =  a.     Consequently 

0  =  F  (a)  +  C, 

whence  C  =  —  F  (a)  and 

A=F{x)-F{a).  I 

This  is  the  area  from  x  =  a  to  the  ordinate  MP  with  abscissa 
x.     The  area  between  x  =  a  and  x  =  b  is  then 

A=F(b)-F(a).  (20b) 

The  difference  F  (b)  —  F  (a)  is  often  represented  by  the 
notation  F  (x)\    ,  that  is, 


F{x) 


=  F(b)-F(a).  (20c) 


21.    Relation  of  the  Definite  and  Indefinite  Integrals.  — ■ 
The  definite  integral    /    /  (x)  dx  is  equal  to  the  area  bounded 

J  a 

by  the  curve  y  =  f  (x),  the  x-axis;  and  the  ordinates  x  =  a, 
x  =  b.     If 

rf  (x)  dx  =  F(x)  +  C, 


- 


by  equation  (20b)    this  area  is  F  (b)  —  F  (a).     We  therefore 
conclude  that 

fbf  (x)  dx  =  F(x)\b  =  F(b)-F  (a),  (21) 

*J  a  I  a 

that  is,  to  find  the  value  of  the  definite  integral     /    /  (x)  dx, 

substitute  x  =  a,  and  x  =  b  in  the  indefinite  integral  J  f  (x)  dx 
and  subtract  the  former  from  the  latter  result. 


Art.  22 


Properties  of  Definite  Integrals 


41 


Example.     Find  the  value  of  the  integral 

11     dx 


£ 


u     1+Z2 


The  value  required  is 
^     dx 


j. 


1  7T 

,  =  tan-1  x     =  tan-1 1  —  tan-1 0  =  -r 
1  +  x2  |  o  4 


22.  Properties  of  Definite  Integrals.  —  A  definite  inte- 
gral has  the  following  simple  properties: 

I.       f*f(x)dx  =  -  faf(x)dx. 
'      II.       f  /  (*)  dx  =  fbf  (x)  dx+  f°f  (v)  dx. 

III.       /    /(a)  dx  =  (b  -  a)f(xd,         a  =  x1  =  b. 

The  first  of  these  is  due  to  the  fact  that  if  Ax  is  positive 
when  x  varies  from  a  to  6,  it  is  negative  when  x  varies  from 
b  to  a.  The  two  integrals  thus  represent  the  same  area  with 
different  algebraic  signs. 


Fig.  22a. 


Fig.  226. 


The  second  property  expresses  that  the  area  from  a  to  c 
is  equal  to  the  sum  of  the  areas  from  a  to  b  and  b  to  c.  This 
is  the  case  not  only  when  b  is  between  a  and  c,  as  in  Fig.  22a, 
but  also  when  b  is  beyond  c,  as  in  Fig.  226.     In  the  latter 

case    /   /  (x)  dx  is  negative  and  the  sum 

fbf(x)dx+  fCf(x)dx 
is  equal  to  the  difference  of  the  two  areas. 


42 


Definite  Integrals 


Chap.  3 


Equation  III  expresses  that  the  area  PQMN  is  equal  to 

that  of  a  rectangle  P'Q'MN  with 
altitude  between  MP  and  NQ. 

23.  Infinite  Limits.  —  It  has 
been  assumed  that  the  limits  a 
and  b  were  finite.     If  the  integral 


Fig.  22c. 


U  a 


dx 


(23) 


approaches  a  limit  when  b  increases  indefinitely,  that  limit 

/  (x)  dx.     That  is, 

a 

£rJo  r*b 

f  (x)  dx  =  lim    /    /  (x)  dx. 
b=oo  nj  a 

If  the  indefinite  integral 

Cf  (x)  dx  =  F  0) 

approaches  a  limit  when  x  increases  indefinitely, 

/    f(x)  dx  =  lim  [F  (b)  -  F  (a)]  =  F  (oo)  -  F  (a). 

*J  a  b  =  oo 

The  value  is  thus  obtained  by  equation  (21)  just  as  if  the 
limits  were  finite. 


Example  1. 


/'•OO 

Jo     I 


dx 


+  x* 


The  indefinite  integral  is 

dx 


h 


+  z2 


=  tan-1  a;. 


TV 


When  x  approaches  infinity,  this  approaches  ^. 


Hence 


J^OO 
o  r 


dx 


=  tan-1  x 


7T 

2 


Art.  24  Infinite  Values  of  the  Function  43 


Ex.  2.       /     cosxdx. 

0 


The  indefinite  integral  sinx  does  not  approach  a  limit 
when  x  increases  indefinitely.     Hence 


/. 


00 

cos  x  dx 
o 


has  no  definite  value. 

24.    Infinite  Values  of  the  Function.  —  If  the  function 

f  (x)   becomes  infinite  when  x  =  b,     /    /  (x)  dx  is  defined  as 

*'  a 

the  limit 

Jf  (x)  dx  —  lim    /    /  (x)  dx, 
a  z  =  b    *J  a 

z  being  between  a  and  b. 
Similarly,  if  /  (a)  is  infinite, 


Jf  (x)  dx  =  lim    /    /  (x)  dx, 
a  z  =  a   *J  z 


z  being  between  a  and  b. 

If  the  function  becomes  infinite  at  a  point  c  between  a  and 

Ch 
b,    I    f  (x)  dx  is  defined  by  the  equation 

fbf(x)dx=    fCf(x)dx+   fbf(x)dx.  (24) 

X1  dx 
~"I7=" 
1  V  X 

When  x  =  0,  -77=    is  infinite.     We  therefore  divide  the 

Vx 

integral  into  two  parts: 


ri  dx^  _    f0dx_,     J 

J_!  \/x        J -i  *Vx        i/C 


^  dx  ?3 

<n ''     2 +  2 : :  a 


44  Definite  Integrals  Chap.  3 


If  we  use  equation  (21),  we  get 

1 

x 


fldx  = 

J -A  X2    ' 


=  -2. 

-i 


Since  the  integral  is  obviously  positive,  the  result  —  2  is 

absurd.     This  is  due  to  the  fact  that  -=  becomes  infinite 

x2 

when  x  =  0.     Resolving  the  integral  into  two  parts,  we  get 
fldx        C°dx  , "   pdx' 

25.  Change  of  Variable.  —  If  a  change  of  variable  is 
made  in  evaluating  an  integral,  the  limits  can  be  replaced  by 
the  corresponding  values  of  the  new  variable.  To  see  this, 
suppose  that  when  x  is  expressed  in  terms  of  t, 


s> 


^  f(x)dx  =  F(x) 
is  changed  into 

4>{t)dt  =  $(«). 


/< 


If  to,  tif  are  the  values  of  t,  corresponding  to  Xo,  Xi, 

F(x0)  =  $(«,        F(x0  =  $(*i),  " 
and  so 

F  (xi)  -  F  (x0)  =  3>  ft)  -  $  (to), 
that  is 


fXlf(x)dx  =    f  \j>(t)dt. 


If  more  than  one  value  of  t  corresponds  to  the  same  value 
of  x,  care  should  be  taken  to  see  that  when  t  varies  from  t0 
to  th  x  varies  from  x0  to  xh  and  that  for  all  intermediate 
values,  /  (x)  dx  =  </»  (t)  dt. 


Example.       I    Va2  —  x2  dx. 

J— a 


Art.  25  Change  of  Variables  45 

Substituting  x  =  a  sin  0,  we  find 

JVtf^tfdx  -  a2  Jcos20(20  =  | [*  +*^|. 

When  x  =  a,  sin  6  =  1,  and  0  =|.     When  x  =  -a,  sin0 


=  -1  and  0  =  —  |.     Therefore 


•n-  * 


Va2-a;2da;  =  a2  /     cos20d0  =  ^ 


0+ Lin  20 


2  Tra2 


IT 

2  2 


Since  sin  f  7r  =  —  1,  it  might  seem  that  we  could  use  fx 
as  the  lower  limit.     We  should  then  get 


a2  r 

*/3tt 


cos2  Odd  =  - 


7r  a2 


2 


This  is  not  correct  because  in  passing  from  |  ir  to  J  tt,  0 
crosses  the  third  and  second  quadrants.  There  cos  0  is 
negative  and 

Va2  —  x2  dx  =  ( —  a  cos  0)  cos  0  c?0, 
and  not  a2  cos2  0  d0  as  assumed  above. 

EXERCISES 
Find  the  values  of  the  following  definite  integrals: 


ir  r6 

1  6.     I    xlnx  dx 

1.    J     setfxdx. 


0 


3 

4  dx 


7-  X^ 

J  a    Va2  —  X2  \ 

2                                                      8.  f     tan 

(x-l)3dx.  4 


3x  +  2 


r3     *rf*    .  9    rl»2U+c-l)dx, 

J--,  v  r»  4-  1 44.  Jo  v 


-5  Vx2  +  144 


.     (**  sin' Odd.  10-    J0 


5 

-x 


1   dx_ 

Vx 


» 


46  Definite  Integrals  Chap.  3 

13.  f     e-*2*2  x  dx. 

14.  f%* 

J  i     x- 

x  Vx2  —  1 


J2 
esc2  x  dx. 
IT 


*  a      C         dx 

.of  dx  14. - 

12.     J 


Evaluate  the  following  definite  integrals  by  making  the  change  of 
variable  indicated: 

1T~\ — ixi'         ^  =  tan  5. 

-i  (1  +  x2)2 

16.     )    d;c,         £  —  1  =  z2. 

•'i         x 


ri        dz  1 

"•  J, 


F|     zV22+l 


X 


7T 


f2  cos  Odd  .    „ 

18.  I       t .  ■        >         sm9=8. 

Jo      6  —  5  sm  6  +  sin2  0 

19.  f°-f-^v  a2+z2=z2. 
Jo    a2  +  x2 


CHAPTER   IV 

SIMPLE   AREAS   AND   VOLUMES 

26.  Area  Bounded  by  a  Plane  Curve.  Rectangular 
Coordinates.  —  The  area  bounded  by  the  curve  y  =f(x), 
the  .r-axis  and  two-  ordinates  x  =  a,  x  =  b,  is  the  limit 
approached  by  the  sum  of  rectangles  y  Ax.     That  is, 


Fig.  26a. 


Fig.  266. 


A  =  lim  x    y  Ax  =    I    y  dx  =   j    f  (x)  dx. 


(26a) 


Similarly,  the  area  bounded  by  a  curve,   the  abscissas 
y  =  a,y  =  b,  and  the  //-axis  is 


A  =  lim  2C  x  Ay  =  I    x  dy. 

Ay=0  J  a 


Example  1.     Find  the  area  bounded  by  the  curve  x 
y  —  if  and  the  {/-axis. 

47 


(26b) 


=  2  + 


48 


Simple  Areas  and  Volumes 


The  curve  (Fig.  26c)  crosses  the  y-axis  at  y 
y  =  2.     The  area  required  is,  therefore, 


Chap.  4 
—  1  and 


Fig.  26c. 


O  N 

Fig.  2Qd. 


A  =  f_xdy  =  J\2  +  y  -  y2)  dy  =  2y  +V~ 


3 


=  4i 


Ex.  2.     Find  the  area  within  the  circle  x2  +  y2  =  16  and 
parabola  x2  =  6  y. 
Solving    the    equations    simultaneously,    we    find    that 

the    parabola   and   circle    intersect   at    P  (—  2V3,  2)    and 

Q  (2 V3,  2) .     The  area  MPQiV  (Fig.  26d)  under  the  circle  is 


'2V3 


•2^3 


/-2V3  f2V3  16 

/        ydx  =  \/16-x2^  =  -^tt+4V3. 

«7-2v/3  «7-2v/3  O 

The  area  MPO  +  OQiV  under  the  parabola  is 

J-2V3  o  3 

The  area  between  the  curves  is  the  difference 
MPQN  -  MPO  -  OQN  =  V5  tt  +  £  V3 . 


Ex.  3.     Find  the  area  within  the  hypocycloid  x  =  a  sin3  <£, 
y  =  a  cos3  <f). 


Art.  26 


Rectangular  Coordinates 


49 


The  area  OAB  in  the  first  quadrant  is 

Jy  dx  =    J    a  cos3  </>  •  3  a  sin2  </>  cos  </>  d<f> 


=  3  a2  ( 

Jo 


»0 


cos4  </>  sin2  </>  c?</>  =  ^\  t  a2. 


Fig.  26e. 

The  entire  area  is  then 

4- OAB  =  |7ra2. 

EXERCISES 

1.  Find  the  area  bounded  by  the  line  2 y  —  3z—  5  =0,  the  x-axis, 
and  the  ordinates  x  —  1,  X  =  3. 

2.  Find  the  area  bounded  by  the  parabola  ?/  =  3  x2,  the  ?/-axis,  and 
the  abscissas  y  —  2,  y  =  4. 

3.  Find  the  area  bounded  by  ?/3  =  x,  the  line  7/  =  —  2,  and  the 
ordinates  x  =  0,  x  =  3. 

4.  Find  the  area  bounded  by  the  parabola  y  =  2  x  —  x-  and  the 
x-axis. 

5.  Find  the  area  bounded  by  y  =  In  x,  the  x-axis,  and  the  ordinates 
x  =  2,  x  =  8. 

6.  Find  the  area  enclosed  by  the  ellipse 

??  4-  yl  =  i 

a'*        //2 

7.  Find  the  area  bounded  by  the  coordinate  axes  and  the  curve 
x*  +  ?/*  =  a*. 


50  Simple  Areas  and  Volumes  Chap.  4 

8.  Find  the  area  within  a  loop  of  the  curve  x2  =  y2  (4  —  y2) . 

9.  Find  the  area  within  the  loop  of  the  curve  y2  =  (x—  1)  (x  —  2)2. 

10.  Show  that  the  area  bounded  by  an  arc  of  the  hyperbola  xy  =  k2, 
the  rc-axis  and  the  ordinates  at  its  ends,  is  equal  to  the  area  bounded  by 
the  same  arc,  the  y-axis  and  the  abscissas  at  its  ends. 

11.  Find  the  area  bounded  by  the  curves  y2  =  4  ax,  x2  =  4  ay.] 

12.  Find  the  area  bounded  by  the  parabola  y  =  2  x  —  x2  and  the 
line  y  =  —  x. 

13.  Find  the  areas  of  the  two  parts  into  which  the  circle  x2  +  y2  =  8 
is  divided  by  the  parabola  y2  =  2  x. 

14.  Find  the  area  within  the  parabola  x2  =  4  y  -f  4  and  the  circle 
x2  -\-  y2  =  16. 

15.  Find  the  area  bounded  by  y2  =  4  x,  x2  =  4  y,  and  x2  +  y2  =  5. 

16.  Find  the '  area  of  a  circle  by  using  the  parametric  equations 
x  =  a  cos  0,  y  =  a  sin  6. 

17.  Find  the  area  bounded  by  the  z-axis  and  one  arch  of  the  cycloid. 

x  =  a  (<£  —  sin  <j>),  y  =  a  (1  —  cos  0). 

18.  Find  the  area  within  the  cardioid 

x  =  a  cos  0(1  —  cos  0),  y  =  a  sin  0  (1  —  cos  0). 

19.  Find  the  area  bounded  by  an  arch  of  the  trochoid, 

x  =  a<$>  —  b  sin  0,         y  =  a  —  h  cos  4>, 

and  the  tangent  at  the  lowest  points  of  the  curve. 

20.  Find  the  area  of  the  ellipse  x2  —  xy  +  y2  =  3. 


Art.- 


xz 


21.  Find  the  area  bounded  by  the  curve  y2  =  77- and  its  as- 

J  y        2a—  x 

ymptote  x  =  2  a. 

22.  Find  the  area  within  the  curve 


x2       /yV 

a2  "*"  W 


1. 


27.  Area  Bounded  by  a  Plane  Curve.  Polar  Coordi- 
nates. —  To  find  the  area  of  the  sector  POQ  bounded  by  two 
radii  OP,  OQ  and  the  arc  PQ  of  a  given  curve. 

Divide  the  angle  POQ  into  any  number  of  equal  parts  A0 
and  construct  the  circular  sectors  shown  in  Fig.  27a.  One 
of  these  sectors  ORS  has  the  area 

i  OR2  ^^  =  i  r2  A0. 


Art.  27 


Polar  Coordinates 


51 


If  a  and  /3  arc  the  limiting  values  of  0,  the  sum  of  all  the 
sectors  is  then 

As  A0  approaches  zero,  this  sum  approaches  the  area  A  of 
the  sector  POQ.     Therefore 


Xp  re 

ir2A6=  / 
a  da. 


1  r2  dd. 


(27) 


Fig.  27a. 


Fig.  276. 


In  this  equation  r  must  be  replaced  by  its  value  in  terms 
of  6  from  the  equation  of  the  curve. 

Example.  Find  the  area  of  one  loop  of  the  curve  r  = 
a  sin  2  0  (Fig.  276). 


A  loop  of  the  curve  extends  from  0  =  0  to  0 


7T 

2* 


Its 


area  is 


7i  n 

J ^2  1  /'2     2 

±r2dd  =    /    ^sin2  (20)  d0 

a2   P 

=H(1- 


cos  4  0)  dd  = 


7TCT 

"8" 


52  Simple  Areas  and  Volumes  Chap.  4 

EXERCISES 

1.  Find  the  area  of  the  circle  r  =  a. 

2.  Find  the  area  of  the  circle  r  =  a  cos  0. 

3.  Find  the  area  bounded  by  the  coordinate  axes  and   the   line 

r  =  a  sec 


(-  f 


4.  Find  the  area  bounded  by  the  initial  line  and  the  first  turn  of  the 
spiral  r  =  ae  • 

5.  Find  the  area  of  one  loop  of  the  curve  r2  =  a2  cos  2  0. 

6.  Find  the  area  enclosed  by  the  curve  r  =  cos  6  +  2. 

7.  Find  the  area  within  the  cardioid  r  =  a  (1  +  cos  0). 

8.  Find  the  area  bounded  by  the  parabola  r  =  a  sec2  -  and  the 
y-axis. 

9.  Find  the  area  bounded  by  the  parabola 

2a 


r  = 


1  —  cos  0 


IT  7T 

and  the  radii  0  —  -r,  0  =  -■ 

4  2 

10.  Find  the  area  bounded  by  the  initial  line  and  the  second  and 
third  turns  of  the  spiral  r  =  ad. 

11.  Find  the  area  of  the  curve  r  =  2  a  cos  3  6  outside  the  circle 
r  =  a. 

12.  Show  that  the  area  of  the  sector  bounded  by  any  two  radii  of 
the  spiral  rd  =  a  is  proportional  to  the  difference  of  those  radii. 

13.  Find  the  area  common  to  the  two  circles  r  =  a  cos  0,  r  = 
a  cos  6  +  a  sin  6. 

14.  Find  the  entire  area  enclosed  by  the  curve  r  =  a  cos3  » • 

15.  Find  the  area  within  the  curve  (r  —  a)2  =  a2  (1  —  6-). 

16.  Through  a  point  within  a  closed  curve  a  chord  is  drawn.  Show 
that,  if  either  of  the  areas  determined  by  the  chord  and  curve  is  a  maxi- 
mum or  minimum,  the  chord  is  bisected  by  the  fixed  point. 

28.  Volume  of  a  Solid  of  Revolution.  —  To  find  the 
volume  generated  by  revolving  the  area  ABCD  about  the 
x-axis. 

Inscribe  in  the  area  a  series  of  rectangles  as  shown  in 
Fig.  28a.     One  of  these  rectangles  PQSR  generates  a  circular 


Art.  23 


Volume  of  a  Solid  of  Revolution 


53 


cylinder  with  radius  y  and  altitude  Ax.     The  volume  of  this 

cylinder  is 

wy2  Ax. 

D 


Fig.  28a. 
If  a  and  b  are  the  limiting  values  of  x,  the  sum  of  the  cylinders 


is 


V  Try2  Ax. 

a 

The  volume  generated  by  the  area  is  the  limit  of  this  sum 

Xb  Pb 

-n-y2  Ax  =    I    Tvy2  dx.  (28) 

If  the  area  does  not  reach  the  axis,  as  in  Fig.  286,  let  i/i 
and  y2  be  the  distances  from  the  axis  to  the  bottom  and  top 

r 


Fig.  236. 

of  the  rectangle   PQRS.     When  revolved  about   the  axis, 
it  generates  a  hollow  cylinder,  or  washer,  of  volume 

7T  (yr  -  2/12)  Ax. 


54  Simple  Areas  and  Volumes  Chap.  4 

The  volume  generated  by  the  area  is  then 

b  f*b 

v  =  lim  V  TV  (y22  -  yx2)  Ax  =    f    tt  (ij22  -  y?)  dx. 

Ax=0  ^*  a  Ua 

If  the  area  is  revolved  about  some  other  axis,  y  in  these 
formulas  must  be  replaced  by  the  perpendicular  from  a  point 
of  the  curve  to  the  axis  and  x  by  the  distance  along  the  axis 
to  that  perpendicular. 

Example  1.  Find  the 
volume  generated  by  re- 
volving the  ellipse 

a2  ^  b2 

O  Ax  ax 

Fig.  28c.  about  the  a>axis. 

From  the  equation  of  the  curve  we  get 

b2 
y2  =  —  (a2  —  x2). 
cr 

The  volume  required  is,  therefore, 

v  =    I     iry2  dx  =  -y   I     (a2  —  x2)  dx  =  -  irdb2. 

J  —  a  Ci      U —a  O 

Ex.  2.     A  circle  of  radius  a  is  revolved  about  an  axis  in  its 
plane  at  the  distance  b 
(greater  than  a)  from  its 
center.    Find  the  volume 
generated. 

Revolve  the  circle, 
Fig.  28d,  about  the  line 
CD.  The  rectangle  MN 
generates  a  washer  with 
radii 

R1=b-x=b-  Va2-y2, 
#2=  6 +£=6+  Va2-y2.  Fig.  2S<2. 

The  volume  of  the  washer  is 

tt  (R22  -  RS)  =  4tt6  Va2  -  if  Ay. 


Art.  28 


Yoi.TTMK    OF    A    SOLID    OF   REVOLUTION 


55 


The  volume  required  is  then 

v=    /  "  4  tt6  Va2  -  if  dy  =  2  w2a2b. 

U —a 

Ex.  3.     Find  the  volume  generated  by  revolving  the  circle 
r  =  a  sin  0  about  the  £-axis. 
In  this  case 

y  =  r  sin  0  =  a  sin2  0, 

x  =  r  cos  0  =  a  cos  0  sin  0. 

The  volume  required  is 


v  =    I  ivy2  dx  =    / 


=    /  Try2  dx  =    /    Tra3  sin4  0  (cos2  0  -  sin2  0)  d0  = 


T2a3 


The  reason  for  using  ir  as  the  lower  limit  and  0  as  the  upper  is 
to  make  dx  positive  along  the  upper  part  ABC  of  the  curve. 


Fig.  28e. 


As  0  varies  from  w  to  0,  the  point  P  describes  the  path 
OABCO.  Along  OA  and  CO  dx  is  negative.  The  integral 
thus  gives  the  volume  generated  by  MABCN  minus  that 
generated  by  OAM  and  OCN. 


EXERCISES 

1.  Find  the  volume  of  a  sphere  by  integration. 

2.  Find  the  volume  of  a  right  cone  by  integration. 

3.  Find  the  volume  generated  by  revolving  about  the  x-axis  the  area 
bounded  by  the  x-axLs  and  the  parabola  y  =  2  x  —  x2. 


56  Simple  Areas  and  Volumes  Chap.   4 

4.  Find  the   volume   generated  by  revolving   about  OY  the  area 

bounded  by  the  coordinate  axes  and  the  parabola  x*  +  y*  =  a5. 

5.  Find  the  volume  generated  by  revolving  about  the  z-axis  the 

area  bounded  by  the  catenary  y  =  »  \e°  + e/ ,  the  z-axis  and  the 

lines  x  —  ±  a. 

6.  Find  the  volume  generated  by  revolving  one  arch  of  the  sine  curve 
y  =  sin  x  about  OX. 

7.  A  cone  has  its  vertex  on  the  surface  of  a  sphere  and  its  axis 
coincides  with  a  diameter  of  the  sphere.     Find  the  common  volume. 

8.  Find  the  volume  generated  by  revolving  about  the  y-sods,  the 
part  of  the  parabola  y2  =  4  ax  cut  off  by  the  line  x  =  a. 

9.  Find  the  volume  generated  by  revolving  about  x  =  a  the  part  of 
the  parabola  y2  =  4  ax  cut  off  by  the  line  re  =  a. 

10.  Find  the  volume  generated  by  revolving  about  y  =  —  2  a  the 
part  of  the  parabola  y2  =  4  ax  cut  off  by  the  line  x  =  a. 

11.  Find  the  volume  generated  by  revolving  one  arch  of  the  cycloid 

x  =  a  (e/>  —  sin  <f>),         y  =  a  (1  —  cos  <f>) 

about  the  a>axis. 

12.  Find  the  volume  generated  by  revolving  the  curve 

x  =  a  cos3  <j>,         y  =  a  sin3  <j> 
about  the  y-axis. 

13.  Find  the  volume  generated  by  revolving  the  cardioid  r  =  a  (1  + 
cos  6)  about  the  initial  line. 

14.  Find   the   volume   generated   by   revolving   the    cardioid   r  = 

a  (1  +  cos  6)  about  the  line  x  =  —  -y 

15.  Find  the  volume  generated  by  revolving  the  ellipse 

x2  +  xy  +  y2  =  3 

about  the  z-axis. 

16.  Find  the  volume  generated  by  revolving  about  the  line  y  =  x 

the  part  of  the  parabola  x2  +  y3  =  a3  cut  off  by  the  line  x  +  y  =  a. 

29.   Volume  of  a  Solid  with  Given  Area  of  Section.  — 

Divide  the  solid  into  slices  by  parallel  planes.  Let  X  be  the 
area  of  section  at  distance  x  from  a  fixed  point.  The  plate 
PQRS  with  lateral  surface  perpendicular  to  PQR  has  the 
volume 

PQR  •  ^x  =  X  Ax. 


Art.  29     Volume  of  a  Solid  with  Given  Area  of  Section 


57 


If  a  and  b  are  the  limiting  values  of  x,  the  sum  of  such  plates  is 

VbXAx. 

The  volume  required  is  the  limit  of  this  sum 

v  =  lim  V^  XAx  =   fbXdx.  (29) 


Ax=0  ~ ^a 


A  j*>--  t»  ckM 


Fig.  29a. 


Example  1.     Find  the  volume  of  the  ellipsoid 

a?      y*      s? 

a2  ^  b-  "*"  c2 


Fig.  296. 

The  section  perpendicular  to  the  z-axis  at  the  distance  x 
from  the  center  is  an  ellipse 

t.  j_  t      i       t. 
62      c2  a2' 


58  Simple  Areas  and  Volumes 

The  semi-axes  of  this  ellipse  are 


Chap.  4 


I 


x 


X1 


MP  =  c\/l-^2,        MQ^b\/l~~ 
By  exercise  6,  page  49,  the  area  of  this  ellipse  is 


7T  •  MP  .  MQ  =  irbc  (l  -  ^\ 


The  volume  of  the  ellipsoid  is,  therefore, 


*J  — a 


/        xL\  4 

be    1 )dx  =  -  wdbc. 

\        a-/  3 


Ex.  2.     The  axes  of  two  jeojial  right  circular  cylinders 
intersect  at  right  angles.     Find  the  common  volume. 


Fig.  29c. 

In  Fig.  29c,  the  axes  of  the  cylinders  are  OX  and  OZ  and 
OABC  is  |  of  the  common  volume.  The  section  of  OABC 
by  a  plane  perpendicular  to  OY  is  a  square  of  side 

MP  =  MQ  =  Va2  -  y2. 


Art.  29      Volume  of  a  Solid  with  Given  Area  of  Section  59 

The  area  of  the  section  is  therefore 

MP  •  M Q  =  a2  -  y2, 

and  the  required  volume  is 

16  a3 


v  =  8  f\a2-  y2)  dy  = 
J  o 


EXERCISES 

1.  Find  the  volume  of  a  pyramid  by  integration. 

2.  A  wedge  is  cut  from  the  base  of  a  right  circular  cylinder  by  a 
plane  passing  through  a  diameter  of  the  base  and  inclined  at  an  angle 
a  to  the  base.     Find  the  volume  of  the  wedge. 

3.  Two  circles  have  a  diameter  in  common  and  lie  in  perpendicular 
planes.  A  square  moves  in  such  a  way  that  its  plane  is  perpendicular 
to  the  common  diameter  and  its  diagonals  are  chords  of  the  circles. 
Find  the  volume  generated. 

4.  The  plane  of  a  moving  circle  is  perpendicular  to  that  of  an  ellipse 
and  the  radius  of  the  circle  is  an  ordinate  of  the  ellipse.  Find  the  vol- 
ume generated  when  the  circle  moves  from  one  vertex  of  the  ellipse  to 
the  other. 

5.  The  plane  of  a  moving  triangle  is  perpendicular  to  a  fixed  diam- 
eter of  a  circle,  its  base  is  a  chord  of  the  circle,  and  its  vertex  lies  on  a 
line  parallel  to  the  fixed  diameter  at  distance  h  from  the  plane  of  the 
circle.  Find  the  volume  generated  by  the  triangle  in  moving  from  one 
end  of  the  diameter  to  the  other. 

6.  A  triangle  of  constant  area  A  rotates  about  a  line  perpendicular 
to  its  plane  while  advancing  along  the  line.  Find  the  volume  swept 
out  in  advancing  a  distance  h. 

7.  Show  that  if  two  solids  are  so  related  that  every  plane  parallel  to 
a  fixed  plane  cuts  from  them  sections  of  equal  area,  the  volumes  of  the 
solids  are  equal. 

8.  A  cylindrical  surface  passes  through  two  great  circles  of  a  sphere 
which  are  at  right  angles.  Find  the  volume  within  the  cylindrical 
surface  and  sphere. 

9.  Two  cylinders  of  equal  altitude  h  have  a  common  upper  base  and 
their  lower  bases  are  tangent.  Find  the  volume  common  to  the  two 
cylinders. 

10.  A  circle  moves  with  its  center  on  the  z-axis  and  its  plane  parallel 
to  a  fixed  plane  inclined  at  45°  to  the  z-axis.  If  the  radius  of  the  circle 
is  always  r  =  Va2  —  z2,  where  z  is  the  coordinate  of  its  center,  find  the 
volume  described. 


CHAPTER  V 

OTHER   GEOMETRICAL  APPLICATIONS 

30.  Infinitesimals  of  Higher  Order.  —  In  the  applica- 
tions of  the  definite  integral  that  we  have  previously  made, 
the  quantity  desired  has  in  each  case  been  a  limit  of  the  form 

lim  2\  f  (x)  Ax. 

Ax=0  ^ a 

We  shall  now  consider  cases  involving  limits  of  the  form 


lim  V  F(x,Ax) 

At=(1     ~4  n 


Ax=0 

when  F  (x,  Ax)  is  only  approximately  expressible  in  the  form 
/  (x)  Ax.  Such  cases  are  usually  handled  by  neglecting 
infinitesimals  of  higher  order  than  Ax.  That  such  neglect 
does  not  change  the  limit  is  indicated  by  the  following 
theorem: 

If  for  values  of  x  between  a  and  b,  F  (x,  Ax)  differs  from 
f  (x)  Ax  by  an  infinitesimal  of  higher  order  than  Ax, 

lim  V6  F  (x,  Ax)  =  lim  V  V  (x)  Ax. 

Az=0   *-' a  Ax=0    ** 'a 

To  show  this  let  €  be  a  number  so  chosen  that 
F  (x,  Ax)  =  /  (x)  Ax  +  e  Ax. 

If  F  (x,  Ax)  and  /  (x)  Ax  differ  by  an  infinitesimal  of  higher 
order  than  Ax,  e  Ax  is  of  higher  order  than  Ax  and  so  e  ap- 
proaches zero  as  Ax  approaches  zero  (Differential  Calculus, 
Art.  9).     The  difference 

V&F(x,  Ax)  -  V7(x)  Ax  -  V*eAx 

60 


< 


oc 


Art.  31 


Rectangular  Coordinates 


61 


is  graphically  represented  by  a  sum  of  rectangles  (Fig.  30), 
whose  altitudes  are  the  various  values  of  e.  Since  all  these 
values  approach  zero  *  with  Az, 
the  total  area  approaches  zero 
and  so 

limV  F(x,Ax)=limV  f(x)Axt 


Y 


Ax 


0 


m 


kn 


-j? 


Fig.  30. 


which  was  to  be  proved. 

31.  Length  of  a  Curve.  Rectangular  Coordinates.  — 
In  the  arc  AB  of  a  curve  inscribe  a  series  of  chords.  The 
length  of  one  of  these  chords  PQ  is 


VAx2  + 


^  =  \ll+ (S)  **> 


Fig.  31a. 


and  the  sum  of  their  lengths  is 


sv.+o- 


The  length  of  the  arc  AB  is  defined  as  the  limit  approached 
by  this  sum  when  the  number  of  chords  is  increased  indefi- 
nitely, their  lengths  approaching  zero. 

*  For  the  discussion  to  be  strictly  accurate  it  must  be  shown  that 
there  is  a  number  larger  than  any  of  the  e's  which  approaches  zero.  In 
the  language  of  higher  mathematics,  the  approach  to  the  limit  must  be 
uniform.  In  ordinary  cases  that  certainly  would  be  true.  A  similar 
remark  applies  to  all  the  applications  of  the  above  theorem. 


• 


e 


62  Other  Geometrical  Applications  Chap. 


The  quantity  Vl+M  is  not  a  function  of  x  alone. 
When  Ax  approaches  zero,  however,  the  difference  of 
yl+    t-     and  y  1  +  (t~)    approaches  zero.     If  then  we 

replace  V  1  "M  \    )    Ax  by  y/l  +  l-^-j    Ax,  the  error  is  an 

infinitesimal  of  higher  order  than  Ax.     Therefore  the  length 
of  arc  is 

dy 
In  applying  this  formula  ~  must  be  determined  from  the 

equation  of  the  curve.     The  result  can  also  be  written 

s=    f    Vdx2  +  dy2.  (31) 

'In  this  formula,  y  may  be  expressed  in  terms  of  x,  or  x  in 

terms  of  y,  or  both  may  be  expressed  in  terms  of  a  parameter. 

In  any  case  the  limits  are  the  values  at  A  and  B  of  the 

variable  that  remains. 

Example  1.     Find  the  length  of  the  arc  of  the  parabola 

y2  =  4  x  between  x  =  0  and  x  =  1. 

clx      y 
In  this  case  -r-  =  §■     The  limiting  values  of  y  are  0  and  2. 

dy      2  &  J 

Hence 


s  = 


f  \1+{iJdy  =  £~\V¥+idy  =  V2  +  ln  (l  +  V2). 


Ex.  2.     Find  the  perimeter  of  the  curve 
x  =  a  cos3  0,     2/  =  a  sin3  0. 

In  this  case 


ds  =  Vdx2  +  dy2  =  V9  a2  cos4  </>  sin2  0+  9  a2  sin4  0  cos2  0  c?0 

=  3  a  cos  0  sin  0  d<fi. 


Art.  32 


Polar  Coordinates 


G3 


One-fourth  of  the  curve  is  described  when  <£  varies  from 


TV 


0  to  ~.     Hence  the  perimeter  is 


IT 

=  4   p3  a 

Jo 


cos  0  sin  <f>  d<f>  =  6  a. 


7. 


EXERCISES 

Find  the  circumference  of  a  circle  by  integration. 
Find  the  length  of  if  =  xz  between  (0,  0)  and  (4,  8). 

Find  the  length  of  x  =  In  sec  y  between  y  =  0  and  y  =  -• 

o 

Find  the  length  of  x  =  \y-  —  \  In  y  between  y  =  1  and  y 
Find  the  length  of  y  =  ex  between  (0,  1)  and  (1,  e). 
Find  the  perimeter  of  the  curve 

2,2         1 
x3  +  y3  =  a3. 

Find  the  length  of  the  catenary 

=  -<;  +  , -9 


=  2. 


2/ 


between  x  =  —  a  and  x  =  a. 

/  8.    Find  the  length  of  one  arch  of  the  cycloid 

x  =  a  (4>  —  sin  </>),         y  =  a  (1  —  cos  0). 

9.  Find  the  length  of  the  involute  of  the  circle 
x  =  a  (cos 6  +  d sin  6),     y  =  a  (sin  d—  6  cos 6), 

between  6  =  0  and  0  =  2  71- . 

10.  Find  the  length  of  an  arc  of  the  cycloid 

x  =  a  {6  +  sin  6),     y  =  a  (1  —  cos  8). 

If  s  is  the  length  of  arc  between  the  origin  and  any  point  (x,  y)  of  the 
same  arch,  show  that  frs* 

s2  =  8  ay. 
■^-  .    - 
32.   Length    of    a    Curve.     Polar    Coordinates.  - 
differential  of  arc  of  a  curve  is  (Differential  Calculus, 
54,  59) 

ds  =  Vdx2  +  dy2  =  Vdr2  +  r2  dd2. 


-The 
Arts. 


64  Other  Geometrical  Applications 

Equation  (31)  is,  therefore,  equivalent  to 

s  = 


=    fDVdr°~ 


+  r2  dd2. 


Chap.  5 


(32) 


Y 

\5 

0 

X 

Fig.  32. 
In  this  case  dr  =  add  and 


In  using  this  formula, 
r  must  be  expressed  in 
terms  of  d  or  6  in  terms 
of  r  from  the  equation 
of  the  curve.  The  limits 
are  the  values  at  A  and  B 
of  the  variable  that  re- 
mains. 

Example.  Find  the 
length  of  the  first  turn 
of  the  spiral  r  =  ad. 


Jo 


2  dd2  +  a262  dd' 


a  /      VT 

Jo 


+  d2dd 


=  ira  Vl  +  4 7T2  +  ^ln  (2.tt  +  Vl  +  4tt2). 

Li 


EXERCISES 

1.  Find  the  circumference  of  the  circle  r  =  a. 

2.  Find  the  circumference  of  the  circle  r  =  2  a  cos  0. 

3.  Find  the  length  of  the  spiral  r  =  eae  between  0  =  0  and  0  =  -• 

Hi 

4.  Find  the  distance  along  the  straight  line  r  =  a  sec  (  0  —  tj  )  from 


T 


^/ 


e  =  o  to  e  =  5 

5.  Find  the  arc  of  the  parabola  r  =  a  sec2  §  0  cut  off  by  the  y-axia. 

6.  Find  the  length  of  one  loop  of  the  curve 

r  =  a  cos4-* 
4 

7.  Find  the  perimeter  of  the  cardioid 

r  =  a  (1  +  cos0). 

a 

8.  Find  the  complete  perimeter  of  the  curve  r  =  a  sin3  ~ . 

o 


Art.  33 


Area  of  a  Surface  of  Revolution 


65 


33.   Area  of  a  Surface  of  Revolution.  —  To  find  the  area 
generated  by  revolving  the  arc  AB  about  the  z-axis. 

Join  A  and  B  by  a  broken  line  with  vertices  on  the  arc. 
Let  x,  y  be  the  coor- 
dinates of  P  and  x  + 
bx,  y  -f-  by  those  of  Q. 
The  chord  PQ  generates 
a  frustum  of  a  cone 
whose  area  is 


it  (2  y  +  by)  PQ  = 
tt(2?/  +  by)  Vbx2+by2. 


o 


&> 


Fig.  33a. 


The  area  generated  by  the  broken  line  is  then 


2)    tt  (2  ?/  +  by)  Vbx*  +  A?/2. 


The  area  £  generated  by  the  arc  J.5  is  the  limit  ap- 
proached by  this  sum  when  bx  and  by  approach  zero.  Neg- 
lecting infinitesimals  of  higher  order,  (2  y  +  by)  vAf  +  A?/2 

can   be   replaced   by  2  y  Vdx'2  +  d#2  =  2y  ds.     Hence   the 
area  generated  is 


S 


-£ 


2  T  y  ds. 


(33a) 


In  this  formula  y  and  ds  must  be  calculated  from  the 
equation  of  the  curve.  The  limits  are  the  values  at  A  and 
B  of  the  variable  in  terms  of  which  they  are  expressed. 

Similarly,  the  area  generated  by  revolving  about  the 
?/-axis  is 


'-£ 


2wxds. 


(33b) 


Example.  Find  the  area  of  the  surface  generated  by 
revolving  about  the  y-ox\s  the  part  of  the  curve  y  =  1  —  x2 
above  the  z-axis. 


66  Other  Geometrical  Applications 

In  this  case 

ds  = 


Chap.  5 


V  1  +  (%Y  dx  =  Vl  +  4;X2  dx. 


\dxj 

The  area  required  is  generated  by  the  part  A B  of  the  curve 
between  x  =  0  and  x  =  1.     Hence 


Fig.  336. 


S  =    f'2Trxds=    J     2irxVl+±x2dx 

O  o         O 


EXERCISES 

1.  Find  the  area  of  the  surface  of  a  sphere. 

2.  Find  the  area  of  the  surface  of  a  right  circular  cone. 

3.  Find  the  area  of  the  spheroid  generated  by  revolving  an  ellipse 
about  its  major  axis. 

2  2  2 

4.  Find  the  area  generated  by  revolving  the   curve  x*  +  if  —  os 
about  the  ?/-axis. 

5.  Find  the  area  generated  by  revolving  about  OX,  the  part  of  the 
catenary 

X 


a 


V  =  2  lC°  +C 


•) 


between  x  =  —  a  and  x  =  a. 

6.    Find  the  area  generated  by  revolving  one  arch  of  the  cycloid 

x  =  a  (cj>  —  sin  </>),         y  —  a  (1  —  cos  4>) 
about  OX. 


Art.  34 


Unconventional  Methods 


G7 


7.  Find  the  area  generated  by  revolving  the  cardioid  r  =  a  (1  -f-  cos  0) 

about  the  initial  lino. 

8.  The  arc  of  the  circle 

x2  +  if  =  a2 

between  (a,  0)  and  (0,  a)  is  revolved  about  the  line  x  +  y  =  a.     Find 
the  area  of  the  surface  generated. 

9.  The  arc  of  the  parabola  y-  =  4  x  between  x  =  0  and  x  =  1  is 
revolved  about  the  line  y  =  —  2.     Find  the  area  generated. 

10.  Find  the  area  of  the  surface  generated  by  revolving  the  lemnis- 


7T 

4' 


cate  r-  =  2  a2  cos  2  0  about  the  line  6  = 

34.  Unconventional  Methods.  —  The  methods  that  have 
been  given  for  finding  lengths,  areas,  and  volumes  are  the 
ones  most  generally  applicable. 
In  particular  cases  other  methods 
may  give  the  results  more  easily. 
To  solve  a  problem  by  integra- 
tion, it  is  merely  necessary  to  ex- 
press the  required  quantity  in  any 
way  as  a  limit  of  the  form  used  in 
defining  the  definite  integral. 

Example  1.  When  a  string  held 
taut  is  unwound  from  a  fixed 
circle,  its  end  describes  a  curve 
called  the  involute  of  the  circle. 
Find  the  length  of  the  part  de- 
scribed when  the  first  turn  of  the 
string  is  unwound. 

Let  the  string  begin  to  unwind  at  A.  When  the  end 
reaches  P  the  part  unwound  QP  is  equal  to  the  arc  AQ. 
Hence 

QP  =  AQ  =  ad. 

When  P  moves  to  R  the  arc  PR  is  approximately  the  arc  of 
a  circle  with  center  at  Q  and  central  angle  A0.     Hence 


Fig.  34a. 


PR  =  ad  Id 


68 


Other  Geometrical  Applications 


Chap.  5 


approximately.     The  length  of  the  curve  described  when  0 
varies  from  0  to  2  w  is  then 


2tt 


s  =  lim  2     ad  A# 


A0=O 


Jr,2ir 


addd  =  2  tcl\ 


<X 


Ex.  2.  Find  the  volume  generated  by  rotating  about  the 
y-SbXis  the  area  bounded  by  the  parabola  x2  =  y  —  1,  the 
x-axis,  and  the  ordinates  x  =  ±1. 

Resolve  the  area  into  slices  by  ordinates  at  distances  Ax 
apart.  When  revolved  about  the  j/-axis,  the  rectangle  PM 
between  the  ordinates  x,  x  +  Ax  generates  a  hollow  cylinder 
whose  volume  is 

7r  (x  +  Ax)2  y  —  it  x2y  =  2  irxy  Ax  +  wy  (Ax)2. 


Fig.  346. 


Fie.  34c. 


Since  vy  (Ax)2  is  an  infinitesimal  of  higher  order  than  Ax, 
the  required  volume  is 

lim  V   2irxyAx=    I    2irx  (1  +  x2)  dx  =  fr. 

Z?x.  3.  Find  the  area  of  the  cylinder  x2  -f-  y2  =  ax  within 
the  sphere  x2  +  y2  +  z2  =  a2. 

Fig.  34c  shows  one-fourth  of  the  required  area.  Divide 
the  circle  OA  into  equal  arcs  As.     The  generators  through 


Art.  34  Unconventional  Methods  C9 

the  points  of  division  cut  the  surface  of  the  cylinder  into 
strips.  Neglecting;  infinitesimals  of  higher  order,  the  area 
of  the  strip  MPQ  is  MP  •  As.  If  r,  0  are  the  polar  coordinates 
of  M,  r  =  a  cos  0  and 

As  =  a  Ad,         MP  =  Vtf  —  r2  =  a  sin  0. 


The  required  area  is  therefore  given  by 

7T 

=   lim  Y2a2sin0A0  =    /    a2sin0c?0. 


to 

1 


Consequently 

£=4a2   J     smddd  =  4a2. 

Jo 

EXERCISES 

1.  Find  the  area  swept  over  by  the  string  in  example  1,  page  67. 

2.  Find  the  area  of  surface  cut  from  a  right  circular  cylinder  by  a 
plane  passing  through  a  diameter  of  the  base  and  inclined  45°  to  the 
base. 

3.  The  axes  of  two  right  circular  cylinders  of  equal  radius  intersect 
at  right  angles.  Find  the  area  of  the  solid  common  to  the  two  cylinders 
(Fig.  29c). 

4.  An  equilateral  triangle  of  side  a  is  revolved  about  a  line  parallel 
to  the  base  at  distance  b  below  the  base.     Find  the  volume  generated. 

5.  The  area  bounded  by  the  h}rperbola  x2  —  y2  =  a2  and  the  lines 
y  =  zta  is  revolved  about  the  z-axis.     Find  the  volume  generated. 

6.  The  vertex  of  a  cone  of  vertical  angle  2  a  is  the  center  of  a  sphere 
of  radius  a.    Find  the  volume  common  to  the  cone  and  sphere. 

7.  The  axis  of  a  cone  of  altitude  h  and  radius  of  base  2  a  is  a  gen- 
erator of  a  cylinder  of  radius  a.  Find  the  area  of  the  surface  of  the 
cylinder  within  the  cone. 

8.  Find  the  area  of  the  surface  of  the  cone  in  Ex.  7  within  the 
cylinder. 

9.  Find  the  volume  of  the  cylinder  in  Ex.  7  within  the  cone. 


CHAPTER  VI  ,. 

MECHANICAL   AND    PHYSICAL   APPLICATIONS 

35.  Pressure.  —  The  pressure  of  a  liquid  upon  a  hori- 
zontal area  is  equal  to  the  weight  of  a  vertical  column  of  the 
liquid  having  the  area  as  base  and  reaching  to  the  surface. 
By  the  pressure  at  a  point  P  in  the  liquid  is  meant  the  pressure 
upon  a  horizontal  surface  of  unit  area  at  that  point.     The 


h 


— > 


i 

Fig.  35a. 


Fig.  356. 


volume  of  a  column  of  unit  section  and  height  h  is  h.     Hence 
the  pressure  at  depth  h  is 

p  =  wh,  (35a) 

w  being  the  weight  of  a  cubic  unit  of  the  liquid. 

To  find  the  pressure  upon  a  vertical  plane  area  (Fig.  356), 
we  make  use  of  the  fact  that  the  pressure  at  a  point  is  the 
same  in  all  directions.  The  pressure  upon  the  strip  AB 
parallel  to  the  surface  is  then  approximately 

pAA, 

p  being  the  pressure  at  any  point  of  the  strip  and  AA  its 
area.     The  reason  for  this  not  being  exact  is  that  the  pressure 

70 


Art.  35  Pressure  71 

at  the  top  of  the  strip  is  a  little  less  than  at  the  bottom. 
This  difference  is,  however,  infinitesimal,  and,  since  it  multi- 
plies AA,  the  error  is  an  infinitesimal  of  higher  order  than 
AA.     The  total  pressure  is,  therefore, 

P  =  lim  V  p  AA  =    f  pdA  =w  fhdA.        (35b) 

Before  integration  dA  must  be  expressed  in  terms  of  h. 
The  limits  are  the  values  of  h  at  the  top  and  bottom  of  the 
submerged  area.  In  case  of  water  the  value  of  w  is  about 
62.5  lbs.  per  cubic  foot. 

Example.     Find  the  water  pressure  upon  a  semicircle  of 


Fig.  35c. 

radius  5  ft.,  if  its  plane  is  vertical  and  its  diameter  in  the 
surface  of  the  water. 

In  this  case  the  element  of  area  is 

dA  =2  V25  -  h2  dh. 
Hence 

P  =  w   fhdA  =  2w  )    h  V25  -  h2  dh 

=  sup.  w  =  2  5  0.  (62#5)  =  5208.3  lbs. 

EXERCISES 

V  1.  Find  the  pressure  sustained  by  a  rectangular  floodgate  10  ft. 
broad  and  12  ft.  deep,  the  upper  edge  being  in  the  surface  of  the  water. 

y  2.  Find  the  pressure  on  the  lower  half  of  the  floodgate  in  the  pre- 
ceding problem. 

••C3.  Find  the  pressure  on  a  triangle  of  base  b  and  altitude  h,  sub- 
merged so  that  its  vertex  is  in  the  surface  of  the  water,  and  its  altitude 
vertical. 

4.  Find  the  pressure  upon  a  triangle  of  base  b  and  altitude  h,  sub- 
merged so  that  its  base  is  in  the  surface  of  the  liquid  and  its  altitude 
vertical. 


72  Mechanical  and  Physical  Applications  Chap.  6 

5.  Find  the  pressure  upon  a  semi-ellipse  submerged  with  one  axis 
in  the  surface  of  the  liquid  and  the  other  vertical. 

I>f6.  A  vertical  masonry  dam  in  the  form  of  a  trapezoid  is  200  ft.  long 
at  the  surface  of  the  water,  150  ft.  long  at  the  bottom,  and  60  ft.  high. 
What  pressure  must  it  withstand? 

7.  One  end  of  a  water  main,  2  ft.  in  diameter,  is  closed  by  a  vertical 
bulkhead.  Find  the  pressure  on  the  bulkhead  if  its  center  is  40  ft. 
below  the  surface  of  the  water. 

8.  A  rectangular  tank  is  filled  with  equal  parts  of  water  and  oil.  If 
the  oil  is  half  as  heavy  as  water,  show  that  the  pressure  on  the  sides  is 
one-fourth  greater  than  it  would  be  if  the  tank  were  filled  with  oil.  ' 

36.  Moment.  —  Divide  a  plane  area  or  length  into  small 
parts  such  that  the  points  of  each  part  differ  only  infinitesi- 
mally  in  distance  from  a  given  axis.  Multiply  each  part  by 
the  distance  of  one  of  its  points  from  the  axis,  the  distance 
being  considered  positive  for  points  on  one  side  of  the  axis 
and  negative  for  points  on  the  other.  The  limit  approached 
by  the  sum  of  these  products  when  the  parts  are  taken 
smaller  and  smaller  is  called  the  moment  of  the  area  or  length 
with  respect  to  the  axis. 

Similarly,  to  find  the  moment  of  a  length,  area,  volume, 
or  mass  in  space  with  respect  to  a  plane,  we  divide  it  into 
elements  whose  points  differ  only  infinitesimally  in  distance 
from  the  plane  and  multiply  each  element  by  the  distance  of 
one  of  its  points  from  the  plane  (considered  positive  for 
points  on  one  side  of  the  plane  and  negative  on  the  other). 
The  moment  with  respect  to  the  plane  is  the  limit  approached 
by  the  sum  of  these  products  when  the  elements  are  taken 
smaller  and  smaller. 

Example.  Find  the  moment  of  a  rectangle  about  an  axis 
parallel  to  one  of  its  sides  at  distance  c. 

Divide  the  rectangle  into  strips  parallel  to  the  axis  (Fig. 
36).  Let  y  be  the  distance  from  the  axis  to  a  strip.  The 
area  of  the  strip  is  b  Ay.     Hence  the  moment  is 

Xc+a  f*c+a  I  a\ 

ybAy  =    I         by  dy  =  able  +  -J. 
c  J  c  \  LI 


1['|\.rt.  37   Center  of  Gravity  of  a  Length  or  Area  in  a  Plane 


73 


[Since  ab  is  the  area  of  the  rectangle  and  c  +  -  is  the  distance 

I  "rom  the  axis  to  its  center,  the  moment  is  equal  to  the  product 

)f  the  area  and  the  distance  from  the  axis  to  the  center  of  the 

cal|rectangle. 

b 


a 


J 


Fig.  36. 


Fig.  37a. 


37.  The  Center  of  Gravity  of  a  Length  or  Area  in  a 
Plane.  —  The  center  of  gravity  of  a  length  or  area  in  a 
plane  is  the  point  at  which  it  could  be  concentrated  without 
changing  its  moment  with  respect  to  any  axis  in  the  plane. 

Let  C  (x,  y)  be  the  center  of  gravity  of  the  arc  AB  (Fig. 
37a),  and  let  s  be  the  length  of  the  arc.  The  moment  of  AB 
with  respect  to  the  z-axis  is 


J  A 


y  ds. 


If  the  length  s  were  concentrated  at  C,  its  moment  would  be 
sy.     By  the  definition  of  center  of  gravity 


sy 


whence 


-f, 

i: 


yds, 


yds 


Similarly, 


x  = 


f 

J  A 


xds 


•C?  TD  . 


74 


Mechanical  and  Physical  Applications 


Chap.  6 


The  limits  are  the  values  at  A  and  B  of  the  variable  in  terms 

of  which  the  integral  is  expressed. 

Let  C  (x,  y)  be  the  center  of  gravity  of  an  area  (Figs.  376, 

37c).     Divide  the  area  into  strips  dA  and  let  (x,  y)  be  the 

center  of  gravity  of  the 
strip  dA .  The  moment 
of  the  area  with  respect 
to  the  x-axis  is 


J  ydA. 


Fig.  376. 


If  the  area  were  con- 
centrated at  C,  the 
moment  would  be  Ay, 
where  A  is  the  total 
area.     Hence 

Ay  =  J  ydA, 


or 


y 

Similarly, 


fydA 


A 


Fig.  37c. 


x  = 


I 


xdA 


A 


The  strip  is  usually  taken  parallel  to  a  coordinate  axis. 
The  area  can,  however,  be  divided  into  strips  of  any  other 
kind  if  convenient. 

Example  1.  Find  the  center  of  gravity  of  a  quadrant  of 
the  circle  x2  +  y2  =  a2. 

In  this  case 

ds  =  Vdx2  +  dy2  =  -  dx 

y 


-     Art.  37    Center  of  Gravity  of  a  Length  or  Area  in  a  Plane        75 

and 

I yds  =    /   Vjdx  =  a2. 

The  length  of  the  arc  is 


Hence 


1   tn       \        w 
s  =  -  (2  tvo)  =  7> a- 


I  y  ds     0 

y  =  — ; —  =  — 


IT 


ria.  o.d. 


It  is  evident  from  the  symmetry  of  the  figure  that  x  has  the 

same  value. 

Ex.  2.  Find  the  center  of  gravity  of  the  area  of  a  semi- 
circle. 

From  symmetry  it  is  evident  that  the  center  of  gravity  is 
in  the  y-axis  (Fig.  37c).  Take  the  element  of  area  parallel 
to  OX.     Then  dA  =  2  x  dy  and 

I  CydA  =  J2xydy  =  2  j\  Va2  -  y2  dy  =  f  a3. 


7T 


The  area  is  A  =  „  «2-     Hence 


y 


-i 


yd  A      4o 


A 


TV 


Fig.  37/. 

Ex.  3.     Find  the  center  of  gravity  of  the  area  bounded  by 
the  x-axis  and  the  parabola  y  =  2x  -  x2. 

Take  the  element  of  area  perpendicular  to  OX.     If  (x,  y) 


76  Mechanical  and  Physical  Applications  Chap.  6 

are  the  coordinates  of  the  top  of  the  strip,  its  center  of  gravity 


i) 


Hence  its  moment  with  respect  to  the  x-axis  is 

^  •  dA  =-f  dx. 
2  2y 


The  moment  of  the  whole  area  about  OX  is  then 


Kdx=£V2 


x  —  x2)2  dx  = 


16 


The  area  is 


15 

A  =    j  y  dx  =    /    (2  x  —  x2)  dx  =  -  • 

Hence  y  =  f.     Similarly, 

fxdA        f(2x2-xs)dx 


x  = 


A 


A 


=  1. 


38.    Center  of  Gravity  of  a  Length,  Area,  Volume,  or 
Mass  in  Space.  —  The  center  of  gravity  is  defined  as  the 

point  at  which  the 
mass,  area,  length,  or 
volume  can  be  con- 
centrated without 
changing  its  moment 
with  respect  to  any 
plane. 

Thus  to  find  the  cen- 
x  ter  of  gravity  of  a  solid 
mass  (Fig.  38a)  cut  it 
into  slices  of  mass  Am. 
If  (x,  y,  z)  is  the  cen- 
ter of  gravity  of  the 
slice,  its  moment  with 
respect  to  the  x?/-plane  is  z  Am  and  the  moment  of  the  whole 
mass  is 


Fig.  38a. 


lim  V  z  Am  =    /  zdm. 


Art,  38  Center  of  Gravity  77 

If  the  whole  mass  M  were  concentrated  at  its  center  of 
gravity  (x,  y,  z),  the  moment  with  respect  to  the  xy-plane 
would  be  zM .     Hence 

I  zdm, 


zM  = 


or 


I  zdm 

z  =  * 


M 


(38) 


Similarly, 


/ x dm  j  ydm 


(38) 


The  mass  of  a  unit  volume  is  called  the  density.     If  then 
dv  is  the  volume  of  the  element  dm  and  p  its  density, 

dm  =  pdv. 

To  find  the  center  of  gravity  of  a  length,  area,  or  volume 
it  is  merely  necessary  to  replace  M  in  these  formulas  by  s, 

S,  or  v. 

Example  1.     Find  the  center  of  gravity  of  the  volume  ot 

an  octant  of  a  sphere  of  radius  a. 
The  volume  of  the  slice  (Fig.  38a)  is 

dv  =  \Trx1dz  =  i7r(a2-z2)<fc. 
Hence 

fzdv  =  £\*(a2  ~  z^zdz  =  TQai' 

The  volume  of  an  octant  of  a  sphere  is  J  ira\     Hence 


/ 


zdv      ^a*      3 


2  =  ■  =   =  ~  CI. 

V  7T      , 

6a 


8 


From  symmetry  it  is  evident  that  x  and  y  have  the  same 
value. 


78 


Mechanical  and  Physical  Applications 


Chap.  6 


Ex.  2.     Find  the  center  of  gravity  of  a  right  circular  cone 
whose  density  is  proportional  to  the  distance  from  its  base. 

Cut  the  cone  into  slices  parallel  to 
the  base.  Let  y  be  the  distance  of  a 
slice  from  the  base.  Except  for  in- 
finitesimals of  higher  order,  its  volume 
h  is  irx2  dy,  and  its  density  is  ky  where 
K  is  constant.     Hence  its  mass  is 

Am  =  kirx2y  dy. 

T 

By    similar    triangles    x  =  ■=-  (h  —  y). 
Hence 


x 


hhrr2  n         N,       ,        k-nrVi3 
(h-y)2ydy  = 


h2 


30 

2h2 


M  =  Jdm  =  Jo   -^  (h  -  y)2ydy  =  — 

Therefore,  finally, 

/  y  dm 


V  = 


M 


2, 
=  -h. 
5 


^ 


EXERCISES 

^\-»  Jkl.   The  wind  produces  a  uniform  pressure  upon  a  rectangular  door. 
Find  the  moment  tending  to  turn  the  door  on  its  hinges. 

2.  Find  the  moment  of  the  pressure  upon  a  rectangular  floodgate 
about  a  horizontal  line  through  its  center,  when  the  water  is  level  with 
the  top  of  the  gate. 

\  3.  A  triangle  of  base  b  and  altitude  h  is  submerged  with  its  base 
horizontal,  altitude  vertical,  and  vertex  c  feet  below  the  surface  of  the 
water.  Find  the  moment  of  the  pressure  upon  the  triangle  about  a 
horizontal  line  through  the  vertex. 

4.    Find  the  center  of  gravity  of  the  area  of  a  triangle. 
v  5.    Find  the  center  of  gravity  of  the  segment  of  the  parabola  y2  =  ax, 
cut  off  by  the  line  x  =  a. 


Art.  38  Center  of  Gravity  79 

6.  Find  the  center  of  gravity  of  the  area  of  a  quadrant  of  the  ellipse 

a2  T  b2 

7.  Find  the  center  of  gravity  of  the  area  bounded  by  the  coordinate 
axes  and  the  parabola  x*  +  y^  =  aK 

8.  Find  the  center  of  gravity  of  the  area  above  the  x-axis  bounded 
by  the  curve  x*  -f-  y*  =  a'. 

9.  Find  the  center  of  gravity  of  the  area  bounded  by  the  x-axis  and 
one  arch  of  the  curve  y  =  sin  x. 

10.  Find  the  center  of  gravity  of  the  area  bounded  by  the  two  parab- 
olas if  =  ax,  x2  =  ay. 

11.  Find  the  center  of  gravity  of  the  area  of  the  upper  half  of  the 
cardioid  r  =  a  (1  +  cos0). 

12.  Find  the  center  of  gravity  of  the  area  bounded  by  the  x-axis  and 
one  arch  of  the  cycloid, 

x  =  a  (0  —  sin <j>),         y  =  a  (1  —  cos  </>). 

13.  Find  the  center  of  gravity  of  the  area  within  a  loop  of  the  lemnis- 
cate  r-  =  a2  cos  2  d. 

14.  Find  the  center  of  gravity  of  the  arc  of  a  semicircle  of  radius  a. 

15.  Find  the  center  of  gravity  of  the  arc  of  the  catenary 

between  x  =  —  a  and  x  =  a. 

16.  Find  the  center  of  gravity  of  the  arc  'of  the  curve  x*  +  y^  =  cfi 
in  the  first  quadrant. 

17.  Find  the  center  of  gravity  of  the  arc  of  the  curve 

x  =  \y2  —  \  In  y 

between  y  —  1  and  y  =  2. 
-\   4   18.    Find  the  center  of  gravity  of  an  arch  of  the  cycloid 

x  =  a  (4>—  sin <£),         y  =  a  (1  —  cos<£). 

19.  Find  the  center  of  gravity  of  a  right  circular  cone  of  constant 
density. 

20.  Find  the  center  of  gravity  of  a  hemisphere  of  constant  density. 

21.  Find  the  center  of  gravity  of  the  solid  generated  by  revolving 
about  OX  the  area  bounded  by  the  parabola  y2  =  4  x  and  the  line  x  =  4. 

22.  Find  the  center  of  gravity  of  a  hemisphere  whose  density  is 
proportional  to  the  distance  from  the  plane  face. 

23.  Find  the  center  of  gravity  of  the  solid  generated  by  rotating  a 
sector  of  a  circle  about  one  of  its  bounding  radii. 


I 


80  Mechanical  and  Physical  Applications  Chap.  6 

24.  Find  the  center  of  gravity  of  the  solid  generated  by  revolving 
the  cardioid  r  =  a  (1  +  cos  6)  about  the  initial  line. 

25.  Find  the  center  of  gravity  of  the  wedge  cat  from  a  right  circular 
cylinder  by  a  plane  passing  through  a  diameter  of  the  base  and  making 
with  the  base  the  angle  «. 

t/26.    Find  the  center  of  gravity  of  a  hemispherical  surface. 
/21.    Show  that  the  center  of  gravity  of  a  zone  of  a  sphere  is  midway 
between  the  bases  of  the  zone. 

28.  The  segment  of  the  parabola  y2  =  2  ax  cut  off  by  the  line  x  =  a 
is  revolved  about  the  z-axis.  Find  the  center  of  gravity  of  the  surface 
generated. 

39.  Theorems  of  Pappus.  Theorem  I.  —  If  the  arc  of  a 
plane  curve  is  revolved  about  an  axis  in  its  plane,  and  not 
crossing  the  arc,  the  area  generated  is  equal  to  the  product 
of  the  length  of  the  arc  and  the  length  of  the  path  described 
by  its  center  of  gravity. 

Theorem  II.  If  a  plane  area  is  revolved  about  an  axis  in 
its  plane  and  not  crossing  the  area,  the  volume  generated  is 
equal  to  the  product  of  the  area  and  the  length  of  the  path 
described  by  its  center  of  gravity. 

To  prove  the  first  theorem,  let  the  arc  be  rotated  about 
the  z-axis.     The  ordinate  of  its  center  of  gravity  is 


/ 


y  ds 
whence 


7r   I  y  ds 


2-wys. 


The  left  side  of  this  equation  represents  the  area  of  the 
surface  generated.  Also  2iry  is  the  length  of  the  path 
described  by  the  center  of  gravity.  This  equation,  therefore, 
expresses  the  result  to  be  proved. 

To  prove  the  second  theorem  let  the  area  be  revolved  about 
the  z-axis.     From  the  equation 


j  ydA 


y 


Art.  39 


Theorems  of  Pappus 


81 


we  get 

2tt  J  ydA  =  2-iryA. 

Since  2  it  j  y  dA  is  the  volume  generated,  this  equation  is 

equivalent  to  theorem  II. 

Example  1.  Find  the  area  of  the  torus  generated  by 
revolving  a  circle  of  radius  a  about  an  axis  in  its  plane  at 
distance  b  (greater  than  a)  from  its 
center. 

Since  the  circumference  of  the  circle 
is  2  ira  and  the  length  of  the  path  de- 
scribed by  its  center  2  irb,  the  area  gen- 
erated is 


S  =  2ira-2irb  =  4:T2ab. 


Fig.  39a. 


Ex.  2.  Find  the  center  of  gravity  of  the  area  of  a  semi- 
circle by  using  Pappus's  theorems. 

When  a  semicircle  of  radius  a  is  revolved  about  its  diameter, 
the  volume  of  the  sphere  generated  is  $  7ra3.  If  y  is  the 
distance  of  the  center  of  gravity  of  the  semicircle  from  this 
diameter,  by  the  second  theorem  of  Pappus, 


whence 


$  ira3  =  2  iry  A  =  2  iry  •  £  ira2, 


y 


7r2a2 


4a 


IT 


Ex.  3.  Find  the  volume  gen- 
erated by  revolving  the  cardioid 
r  =  a  (1  +  cos  6)  about  the  initial 
line. 

The  area  of  the  triangle  OPQ 
is  approximately 
Fig.  396.  i  r2  A0, 

and  its  center  of  gravity  is  §  of  the  distance  from  the  vertex 

to  the  base.     Hence 

y  =  |  r  sin0. 


82  Mechanical  and  Physical  Applications  Chap.  6 

By  the  second  theorem  of  Pappus,  the  volume  generated  by 
OPQ  is  then  approximately 

2  iry  A  A  =  §  tit3  sin  0  A0. 
The  entire  volume  is  therefore 

v  =    /§7rr3sin0d0  =  f7ra3  /     (1  +  cos0)3sin0d0 

«/0  i/0 


2      ,(l  +  cos0)4|*-      8 


? 


EXERCISES 

1.  By  using  Pappus's  theorems  find  the  lateral  area  and  the  volume 
of  a  right  circular  cone. 

2.  Find  the  volume  of  the  torus  generated  by  revolving  a  circle  of 
radius  a  about  an  axis  in  its  plane  at  distance  b  (greater  than  a)  from 
its  center. 

3.  A  groove  with  cross-section  an  equilateral  triangle  of  side  \  inch 
is  cut  around  a  cylindrical  shaft  6  inches  in  diameter.  Find  the  volume 
of  material  cut  away. 

ir  4.  A  steel  band  is  placed  around  a  cylindrical  boiler  48  inches  in 
diameter.     A  cross-section  of  the  band  is  a  semi-ellipse,  its  axes  being  6 

and  V6  inches,  respectively,  the  greater  being  parallel  to  the  axis  of  the 
boiler.     What  is  the  volume  of  the  band? 
^  5.    The  length  of  an  arch  of  the  cycloid 

x  —  a  (<f>  —  sin  0),         y  =  a  (1  —  cos  <f>) 

is  8  a,  and  the  area  generated  by  revolving  it  about  the  x-axis  is  -^  ira2. 
Find  the  area  generated  by  revolving  the  arch  about  the  tangent  at  its 
highest  point. 

6.  By  the  method  of  Ex.  3,  page  81,  find  the  volume  generated  by 
revolving  the  lemniscate  r2  =  2  a2  cos  2  d  about  the  x-axis. 

7.  Obtain  a  formula  for  the  volume  generated  by  revolving  the 
polar  element  of  area  about  the  line  x  =  —  a.  Apply  this  formula  to 
obtain  the  volume  generated  by  revolving  about  x  =  —  a  the  sector  of 
the  circle  r  =  a  bounded  by  the  radii  6  =  —  a,  6  =  +  a. 

8.  A  variable  circle  revolves  about  an  axis  in  its  plane.  If  the 
distance  from  the  center  of  the  circle  to  the  axis  is  2  a  and  its  radius 
is  a  sin  8,  where  9  is  the  angle  of  rotation,  find  the  volume  of  the  horn- 
shaped  solid  that  is  generated. 

9.  Can  the  area  of  the  surface  in  Ex.  8  be  found  in  a  similar  way? 


Art.  40 


Moment  of  Inertia 


83 


10.  The  vertex  of  a  right  circular  cone  is  on  the  surface  of  a  right 
circular  cylinder  and  its  axis  cuts  the  axis  of  the  cylinder  at  right  angles. 
Find  the  volume  common  to  the  cylinder  and  cone  (use  sections  deter- 
mined by  planes  through  the  vertex  of  the  cone  and  the  generators  of 
the  cylinder). 

40.  Moment  of  Inertia.  —  The  moment  of  inertia  of  a 
particle  about  an  axis  is  the  product  of  its  mass  and  the  square 
of  its  distance  from  the  axis. 

To  find  the  moment  of  inertia  of  a  continuous  mass,  we 
divide  it  into  parts  such  that  the  points  of  each  differ  only 
infinitesimally  in  distance  from  the  axis.  Let  Am  be  such  a 
part  and  R  the  distance  of  one  of  its  points  from  the  axis. 
Except  for  infinitesimals  of  higher  order,  the  moment  of 
inertia  of  Am  about  the  axis  is  R2  Am.  The  moment  of 
inertia  of  the  entire  mass  is  therefore 


I  =  Urn  y.R2Am  =    fR2dm. 


(40) 


By  the  moment  of  inertia  of  a  length,  area,  or  volume,  we 
mean  the  value  obtained  by  using  the  differential  of  length, 
area,  or  volume  in  place  of 
dm  in  equation  (40). 

Example  1.  Find  the  mo- 
ment of  inertia  of  a  right  cir- 
cular cone  of  constant  density 
about  its  axis. 

Let  p  be  the  density,  h  the 
altitude,  and  a  the  radius  of 
the  base  of  the  cone.  Divide 
it  into  hollow  cylindrical  slices 
by  means  of  cylindrical  sur- 
faces having  the  same  axis  as 
the  cone.  By  similar  triangles 
the  altitude  y  of  the  cylin- 
drical surface  of  radius  r  is 

y  —  —  {a 


Fig.  40a. 


r). 


84 


Mechanical  and  Physical  Applications 


Chap.  6 


Neglecting  infinitesimals  of  higher  order,  the  volume  between 
the  cylinders  of  radii  r  and  r  -f-  Ar  is  then 

2tt/i 


Av  =  2  irry  Ar 


a 


r  (a  —  r)  dr. 


The  moment  of  inertia  is  therefore 

2irhp 


I  =    I  r2  dm  =    I 


dm  =    I    r2p  dv  = 


a 


f 

I/O 


r3  (a  —  r)  dr  = 


irpha4 


The  mass  of  the  cone  is 


Y 

ft  .'■'.■ 

(         V 

x      \ 

V      ° 

J          r 

Hence 


M  =  pv  =  lirpa2h. 


I  = 


3 
TO 


Ma2. 


Ex.  2.  Find  the  moment  of  in- 
ertia of  the  area  of  a  circle  about 
a  diameter  of  the  circle. 

Let  the  radius  be  a  and  let  the  re- 
axis  be  the  diameter  about  which 
the  moment  of  inertia  is  taken. 
Divide  the  area  into  strips  by  lines  parallel  to  the  x-axis. 
Neglecting  infinitesimals  of  higher  order,  the  area  of  such  a 
strip  is  2  x  Ay  and  its  moment  of  inertia  2  xy2  Ay.  The 
moment  of  inertia  of  the  entire  area  is  therefore 


Fig.  406. 


I  =  J2xy2dy  =  2  J_Va-  -  fy2dy  = 


Traq 


EXERCISES 

«M..    Find  the  moment  of  inertia  of  the  area  of  a  rectangle  about  one 
of  its  edges. 

2.  Find  the  moment  of  inertia  of  a  triangle  about  its  base. 

3.  Find  the  moment  of  inertia  of  a  triangle  about  an  axis  through 
its  vertex  parallel  to  its  base. 

•  4.    Find  the  moment  of  inertia  about  the  ?/-axis  of  the  area  bounded 
by  the  parabola  y2  =  4  ax  and  the  line  x  =  a. 

5.  Find  the  moment  of  inertia  of  the  area  in  Ex.  4  about  the  line 
x  =  a. 

6.  Find  the  moment  of  inertia  of  the  area  of  a  circle  about  the  axis 
perpendicular  to  its  plane  at  the  center.  (Divide  the  area  into  rings 
with  centers  at  the  center  of  the  circle.) 


Art.  41  Work  Done  by  a  Force  85 

7.  Find  the  moment  of  inertia  of  a  cylinder  of  mass  M  and  radius  a 
about  its  axis. 

8.  Find  the  moment  of  inertia  of  a  sphere  of  mass  M  and  radius  a 
about  a  diameter. 

•'S.    An  ellipsoid  is  generated  by  revolving  the  ellipse 

a2  T  ¥ 

about  the  x-axis.     Find  its  moment  of  inertia  about  the  x-axis. 

10.  Find  the  moment  of  inertia  of  a  hemispherical  shell  of  constant 
density  about  the  diameter  perpendicular  to  its  plane  face. 
^"11.  Prove  that  the  moment  of  inertia  about  any  axis  is  equal  to  the 
moment  of  inertia  about  a  parallel  axis  through  the  center  of  gravity 
plus  the  product  of  the  mass  and  the  square  of  the  distance  between  the 
two  axes. 

12.  Use  the  answer  to  Ex.  6,  and  the  theorem  of  Ex.  11  to  determine 
the  moment  of  inertia  of  a  circular  area  about  an  axis,  perpendicular  to 
its  plane  at  a  point  of  the  circumference. 

41.  Work  Done  by  a  Force.  —  Let  a  force  be  applied  to 
a  body  at  a  fixed  point.  When  the  body  moves  work  is  done 
by  the  force.  If  the  force  is  constant,  the  work  is  denned  as 
the  product  of  the  force  and  the  distance  the  point  of  appli- 
cation moves  in  the  direction  of  the  force.     That  is, 

I  W  =  Fs,  (41a) 

where  W  is  the  work,  F  the  force,  and  s  the  distance  moved 
in  the  direction  of  the  force. 

If   the   direction   of    motion    does 
not  coincide  with  that  of  the  force, 

the  work  done  is  the  product  of  the   a^^6 I  >-F 

force  and  the  projection   of  the  dis-  Fig.  41a. 

placement  on  the  force.     Thus  when  the  body  moves  from 
A  to  B  (Fig.  41a)  the  work  done  by  the  force  F  is 

[  W  =  Fs  cos  0.  (41b) 

If  the  force  is  variable,  we  divide  the  path  into  parts  As. 
In  moving  the  distance  As,  the  force  is  nearly  constant  and 
so   the   work   done   is   approximately   F  cos  0  As.     As   the 


86 


Mechanical  and  Physical  Applications 


Chap.  6 


intervals  As  are  taken  shorter  and  shorter,  this  approximation 
becomes  more  and  more  accurate.  The  exact  work  is  then 
the  limit 


W  =  lim  V  F  cos  6  As  =    /  F  cos  6  ds. 

As=0  ^  J 


(41c) 


To  determine  the  value  of  W,  we  express  F  cos  6  and  ds 
in  terms  of  a  single  variable.  The  limits  of  integration  are 
the  values  of  this  variable  at  the  two  ends  of  the  path.  If 
the  displacement  is  in  the  direction  of  the  force,  6  =  0, 
cos  0  =  1  and 


/ 


W  =    /  Fds. 


(41d) 


*B 


Fig.  416. 


Fig.  41c. 


Example  1.  The  amount  a  helical  spring  is  stretched  is 
proportional  to  the  force  applied.  If  a  force  of  100  lbs.  is 
required  to  stretch  the  spring  1  inch,  find  the  work  done  in 
stretching  it  4  inches. 

Let  s  be  the  number  of  inches  the  spring  is  stretched.  The 
force  is  then 

F  =  ks, 

k  being  constant.     When  s  =  1,  F  =  100  lbs.     Hence  k  = 
100  and 

F  =  100  s. 

The  work  done  in  stretching  the  spring  4  inches  is 

JFds  =    I    100  s  ds  =  800  inch  pounds  =  66 J  foot  pounds, 
o  Jo 


I 


Art.  41 


Work  Done  by  a  Force 


87 


Ex.  2.  A  gas  is  confined  in  a  cylinder  with  a  movable  piston. 
Assuming  Boyle's  law  pv  =  k,  find  the  work  done  by  the 
pressure  of  the  gas  in  pushing  out  the  piston  (Fig.  41d). 

Let  v  be  the  volume  of  gas  in  the  cylinder  and  p  the  pressure 
per  unit  area  of  the  piston.  If  A  is  the  area  of  the  piston, 
pA  is  the  total  pressure  of  the  gas  upon  it.  If  s  is  the  distance 
the  piston  moves,  the  work  done  is 

pA  ds. 
But  A  ds  =  dv.     Hence 


II   =    /     pdu  =    /     -  dv  = 


fcln?* 

Vi 


is  the  work  done  when  the  volume  expands  from  V\  to  v2. 


— s- 


' ,  »    . 


-   i      y  , 


-    ,  .   '     '  •   *  / 


Fig.  Aid. 


a  A 

Fig.  41e. 


Ex.  3.     The  force  with  which  an  electric  charge  ei  repels 
a  charge  e2  at  distance  r  is 


where  k  is  constant.  Find  the  work  done  by  this  force 
when  the  charge  e2  moves  from  r  =  a  to  r  =  b,  cx  remaining 
fixed.  V 

Let  the  charge  c2  move  from  A  *%q^B  along  any  path  AB 
(Fig.  41e).     The  work  done  by  the  force  of  repulsion  is 

l.<?2 


w 


dr 


=  jFcosdds=  fFdr=   f*5^ 

The  work  depends  only  on  the  end  points  A  and  B  and  not 
on  the  path  connecting  them. 


88 


Mechanical  and  Physical  Applications 


Chap.  6 


EXERCISES 

1.  According  to  Hooke's  law  the  force  required  to  stretch  a  bar  from 
the  length  a  to  the  length  a  +  x  is 

kx 
a 

where  K  is  constant.     Find  the  work  done  in  stretching  the  bar  from  the 
length  a  to  the  length  b. 

2.  Supposing  the  force  of  gravity  to  vary  inversely  as  the  square  of 
the  distance  from  the  earth's  center,  find  the  work  done  by  gravity  on 
a  meteor  of  weight  w  lbs.,  when  it  comes  from  an  indefinitely  great 
distance  to  the  earth's  surface. 

3.  If  a  gas  expands  without  change  of 
temperature,  according  to  van  der  Waal's 
equation, 


V  = 


a 


v  —  b 


-,2> 


a,  b,  c  being  constant.  Find  the  work  done 
when  the  gas  expands  from  the  volume  vx  to 
the  volume  v2. 

4.   The  work  in   foot   pounds   required 
to    move    a    body   from    one    altitude  to 
another   is    equal   to    the   product    of   its 
weight  in  pounds  and  the  height   in  feet 
IG*      ■''  that  it  is  raised.     Find  the  work  required 

to  pump  the  water  out  of  a  cylindrical  cistern  of  diameter  4  ft.  and 

depth  8  ft. 

5.  A  vertical  shaft  is  supported  by  a  flat  step  bearing  (Fig.  41/). 
The  frictional  force  between  a  small  part  of  the  shaft  and  the  bearing  is 
fxp,  where  p  is  the  pressure  between  the  two  and  /x  is  a  constant.  If  the 
pressure  per  unit  area  is  the  same  at  all  points  of  the  supporting  surface, 
and  the  weight  of  the  shaft  and  its  load  is  P,  find  the  work  of  the  fric- 
tional forces  during  each  revolution  of  the  shaft. 

6.  When  an  electric  current  flows  a  distance  x  through  a  homo- 
geneous conductor  of  cross-section  A,  the  resistance  is 

kx 
A' 

where  K  is  a  constant  depending  on  the  material.  Find  the  resistance 
when  the  current  flows  from  the  inner  to  the  outer  surface  of  a  hollow 
cylinder,  the  two  radii  being  a  and  6. 


Art.  41 


Work  Done  by  a  Force 


89 


7.  Find  the  resistance  when  the  current  flows  from  the  inner  to  the 
outer  surface  of  a  hollow  sphere. 

8.  Find  the  resistance  when  the  current 
flows  from  one  base  of  a  truncated  cone  to 
the  other. 

.  9.  When  an  electric  current  i  flows  an  in- 
finitesimal distance  AB  (Fig.  4lg)  it  produces 
at  any  point  0  a  magnetic  force  (perpendicular 
to  the  paper)  equal  to 

idO 
r   '  Fig.  41<7. 

where  r  is  the  distance  between  AB  and  0.     Find  the  force  at  the  center 
of  a  circle  due  to  a  current  i  flowing  around  it. 

10.    Find  the  magnetic  force  at  the  distance  c  from  an  infinite  straight 
line  along  which  a  current  i  is  flowing. 


CHAPTER   VII 


APPROXIMATE    METHODS 

42.   The    Prismoidal    Formula.  —  Let    ylf    y3,    be    two 

ordinates  of  a  curve  at  distance  h  apart,  and  let  y2  be  the 

ordinate  midway  between 
them.  The  area  bounded 
by  the  z-axis,  the  curve,  and 
the  two  ordinates  is  given 
approximately  by  the  for- 
mula 

Fig.  42a.  A  =  ih  (Vi+  42/2+  2fe).    (42a) 

This  is  called  the  prismoidal  formula  because  of  its  similarity 
to  the  formula  for  the  volume  of  a  prismoid. 
If  the  equation  of  the  curve  is 


y  =  a  +  bx  +  ex2  +  dx3, 


(42b) 


where  a,  b,  c,  d,  are  constants  (some  of  which  may  be  zero), 
the  prismoidal  formula  gives  the  exact  area.  To  prove  this 
let  k  be  the  abscissa  of  the  middle  ordinate  and  t  the  dis- 
tance of  any  other  ordinate  from  it  (Fig.  42a).     Then 

x  =  k  +  t. 

If  we  substitute  this  value  for  x,  (42b)  takes  the  form 

y  =  A  +  Bt  +  Ct2  +  Dt3, 

where  A,  B,  C,  D  are  constants.     The  ordinates  ?/i,  y2,  yz  are 

■    h       h 
obtained  by  substituting  t  =  —  x,  0,  ^.     Hence 

2/i  +  42/2  +  2/3  =  6  A+  i  Ch*. 
90 


Art.  42 

The  Prismoidal  Formula 

Also  the  area 

is 

A 

-2 

=  Ah  +  C 

/i3 
12* 

This  is 

equivalent  to 

h 
6 

U  A  +  1  C/i2) 

-g<*  + 

4  2/2 

H-2/s), 

91 


which  was  to  be  proved. 

If  the  equation  of  the  curve  does  not  have  the  form  (42b), 
it  may  be  approximately  equivalent  to  one  of  that  type  and 
so  the  prismoidal  formula  may  give  an  approximate  value 
for  the  area. 

While  we  have  illustrated  the  prismoidal  formula  by  the 
area  under  a  curve,  it  may  be  used  equally  well  to  determine 
a  length  or  volume  or  any  other  quantity  represented  by  a 
definite  integral, 


r. 


f  (x)  dx. 


Since  such  an  integral  represents  the  area  under  the  curve 
V  =  /  0*0  >  its  value  can  be  found  by  replacing  h  in  (42a)  by 


b  -  a  and  yu  y2l  y3  by  /  (a),  / 

Example  1.  Find  the 
area  bounded  by  the 
a:-axis,  the  curve  y  = 
e~z~,  and  the  ordinates 
x  =  0,  x  =  2. 

The  integral 


'a  +  V 


,  f  (b)  respectively. 


/• 


-l2 


dx 


Fig.  426. 


cannot    be    expressed    in    terms    of   elementary   functions. 
Therefore  we  cannot  obtain  the  area  by  the  methods  that  we 


92 


Approximate  Methods 


Chap.  7 


have  previously  used.  The  ordinates  2/1,  y2)  yz,  in  this  case 
are 

2/i  =  l,         2/2  =  e-1,         2/3  =  e-2- 

The  prismoidal  formula,  therefore,  gives 

The  answer  correct  to  3  decimals  (obtained  from  a  table)  is 
0.882. 

Ex.  2.  Find  the  length  of  the  parabola  y2  =  4  x  from 
x  =  1  to  x  =  5. 

The  length  is  given  by  the  formula 


+  -  +  -)  =  0.869. 
e      el 


-/w 


cfa. 


By  integration  we  find  s  =  4.726.     To  apply  the  prismoidal 
formula,  let 


y 


-^ 


Then  /i  =  4, 


and 


2/i 


=  V2 


2/2 


-VJ, 


2/3 


=  \/f 


5» 


s  =  *(V2  +  4V|  +  Vf)=  4.752. 


Fig.  42c. 
spheroid  perpendicular  to  OX  has  the  area 

A-m'— w(i-S)- 


£x.  3.  Find  the  vol- 
ume of  the  spheroid 
generated  by  revolving 
the  ellipse 

xr      yr       1 

a- "*"  62  " 

about  the  x-axis. 

The    section     of     the 


Art.  43 


Simpson's  Rule 


93 


Its  volume  is 


%J  —  a 


A  dx. 


Since  A  is  a  polynomial  of  the  second  degree  in  x  (a  special 
case  of  a  third  degree  polynomial),  the  prismoidal  formula 
gives  the  exact  volume.  The  three  cross-sections  corre- 
sponding to  x  =  —  a,  x  =  0,  x  =  a,  are 

Ai  =  0,        A2  =  tt62,        Az  =  0. 
Hence 

V  =  ^[Al  +  4A2  +  A3]=^Tab\ 

43.  Simpson's  Rule.  —  Divide  the  area  between  a  curve 
and  the  a>axis  into  any  even  number  of  parts  by  means  of 
equidistant  ordinates  j/i,  y2,  2/3>  ...  ,  yn-  (An  odd  number 
of  ordinates  will  be  needed.)  Simpson's  rule  for  determining 
approximately  the  area  between  yL  and  yn  is 

%  +  4  2/2  +  2  ys  +  47/4  +  21/5+  ■  •  •  +  yn\ 


A  =  h 


(43) 


1+4  +  2  +  4  +  2+   •  •  •  +1 

h  being  the  distance  between  the  ordinates  yi  and  yn.  In 
the  numerator  the  end  coefficients  are  1.  The  others  are 
alternately  4  and  2.  The 
denominator  is  the  sum  of 
the  coefficients  in  the  num- 
erator. 

This  formula  is  obtained 
by  applying  the  prismoidal 
formula  to  the  strips  taken 
two  at  a  time  and  adding 
the  results.    Thus  if  the  area 


Fig.  43. 


is  divided  into  four  strips  by  the  ordinates  ylt  y2,  2/3,  y±,  2/5, 

the  part  between  ?/i  and  y3  has  a  base  equal   to  -x.      Its 
area  as  given  by  the  prismoidal  formula  is 

7T   o(t/l  +4?/2  +  2/3). 


94  Approximate  Methods  Chap.  7 

Similarly  the  area  between  yz  and  2/5  is 

q  2  (2/3 +  4  2/4  +  2/5). 

The  sum  of  the  two  is 

A  =  h  IVi  +  4  ?/2  +  2  y3  +  4  ?/4  +  Vo\ 

By  using  a  sufficiently  large  number  of  ordinates  in 
Simpson's  formula,  the  result  can  be  made  as  accurate  as 
desired. 

Example.     Find  In  5  by  Simpson's  rule.     Since 


ndx 
Ji    x 


in  5  = 

we  take  y  =  -  in  Simpson's  formula.     Dividing  the  interval 
into  4  parts  we  get 

In  5  =  4  (l±M±M±ii±i)  =  L622. 

If  we  divide  the  interval  into  8  parts,  we  get 

In  5  =  &(l  +  t  +  t  +  t  +  t  +  f+i  '+!  +  *)  =  1.6108. 

The  value  correct  to  4  decimals  is 

In  5  =  1.6094. 

44.  Integration  in  Series.  —  In  calculating  integrals  it 
is  sometimes  convenient  to  expand  a  function  in  infinite 
series  and  then  integrate  the  series.  This  is  particularly  the 
case  when  the  integral  contains  constants  for  which  numerical 
values  are  not  assigned.  For  the  process  to  be  valid  all 
series  used  should  converge. 

Example.     Find  the  length  of  a  quadrant  of  the  ellipse 

a2  "*"  62 


Art.  44  Integration  in  Series  95 

Let  a  be  greater  than  b.     Introduce  a  parameter  <f>  by  the 

equation 

x  =  a  sin  (/>. 

Substituting  this  value  in  the  equation  of  the  ellipse,  we  find 

y  =  b  cos  4>. 

Using  these  values  of  x  and  y  we  get 

s  =   fVdx2  +  dif  =   f  Va2  -  (a2  -  62)  sin2  <t>  d4>. 

This  is  an  elliptic  integral.  It  cannot  be  represented  by  an 
expression  containing  only  a  finite  number  of  elementary 
functions.  We  therefore  express  it  as  an  infinite  series.  By 
the  binomial  theorem 

Vd1  -  (a-  -  b2)  sin2  0 

[I    a1  —   ft*  I      /a2  _   £2\2  "1 

1  -  ^  -^-an**-  2l(-^-j  rin«*  .  .  .J- 

Since 

sin2  (f>d(f)  =  -:)  I     snr  <f>d(f)  =  —  > 

o  4  t/o  t         It) 

we  find  by  integrating  term  by  term 

S_a[_2      8       a2  128V     a2     /      s    J 

^Traf        a2-fr2       3  (a2  -  b2V  1 

"  2  |_  4a2         64\     a2     /      '     J" 

If  a  and  b  are  nearly  equal,  the  value  of  s  can  be  calculated 
very  rapidly  from  the  series., 

EXERCISES 

1.  Show  that  the  prismoidal  formula  gives  the  correct  volume  in 
each  of  the  following  cases:  (a)  sphere,  (6)  cone,  (r)  cylinder,  (d) 
pyramid,  (e)  segment  of  a  sphere,  (/)  truncated  cone  or  pyramid. 

2.  Find  the  error  when  the  value  of  the  integral  I  xAdx  is  found 
by  the  prismoidal  formula. 


96  Approximate  Methods  Chap.  7 

In  each  of  the  following  cases  compare  the  value  given  by  the  pris- 
moidal  formula  with  the  exact  value  determined  by  integration. 

3.  Area  bounded  by  y  =  Vx,     y  =  0,     x  =  1,     x  —  3. 

4.  Arc  of  the  curve  y  =  x3  between  x  =  —  2,  x  =  +  2. 

5.  Volume  generated  by  revolving  about  OX  one  arch  of  the  sine 
curve  y  =  sin  x. 

6.  Area  of  the  surface  of  a  hemisphere. 

Compute  each  of  the  following  by  Simpson's  rule  using  4  intervals: 

4  ~  Jo    1  +  x2' 

8.   r9   dx 

Ji  Vl+x3 
9.   Length  of  the  curve  y  =  \nx  from  x  =  1  to  x  =  5. 

10.  Surface  of  the  spheroid  generated  by  rotating  the  ellipse  x2  + 
4  y2  =  4  about  the  x-axis. 

11.  Volume  of  the  solid  generated  by  revolving  about  the  x-axis  the 

j_ 

1  +x2 


area  bounded  by  y  —  0,  y  =  - — ; — - ,  x  =  —  2,     x  =  2. 


12.    Find  the  value  of 


by  expanding  in  series. 
13.    Express 


J    cos  (x2)  dx. 
0 


X 


°  sin  (\x)  dx 


x 


as  a  series  in  powers  of  X. 

14.    Find  the  length  of  a  quadrant  of  the  ellipse  x2  +  2y2  =  2. 


CHAPTER  VIII 
DOUBLE   INTEGRATION 

45.   Double  Integrals.  — ■  The  notation 

/      /    f(x,y)dxdy 

%J  a     *J  c 

is  used  to  represent  the  result  of  integrating  first  with  respect 
to  y  (leaving  x  constant)  between  the  limits  c,  d  and  then 
with  respect  to  x  between  the  limits  a,  b. 

As  here  defined  the  first  integration  is  with  respect  to  the 
variable  whose  differential  stands  last  and  its  limits  are 
attached  to  the  last  integral  sign.  Some  writers  integrate 
in  a  different  order.  In  reading  an  article  it  is  therefore 
necessary  to  know  what  convention  the  author  uses. 

Example.     Find  the  value  of  the  double  integral 


Jo   J —. 


(x2  +  y2)  dx  dy. 


We  integrate  first  with  respect  to  y  between  the  limits  —x, 
x,  then  with  respect  to  x  between  the  limits  0,  1.  The  result 
is 

/     /    (x2  +y2)  dx  dy  =  Jq  dx  (x2y  +  J  y*)x_x  =  J    §  x3  dx  =  f . 

46.  Area  as  a  Double  Integral.  —  Divide  the  area  be- 
tween two  curves  y  =  f  (x),  y  =  F  (x)  into  strips  of  width 
Ax.  Let  P  be  the  point  (x,  y)  and  Q  the  point  (x  +  Ax,  y  + 
Ay).  The  area  of  the  rectangle  PQ  is  Ax  Ay.  The  area 
of  the  rectangle  RS  (Fig.  46a)  is 

XF(x)  PF{x) 

Ay  =  Ax    I        dy. 
m    J         J  ax)    y 

97 


98  Double  Integration  Chap.  8 

The  area  bounded  by  the  ordinates  x  =  a,  x  =  b  is  then 

Xb  f*F{x)  f*b     PF(x) 

Ax  I        dy  =    I      J        dx  dy. 

If  it  is  simpler  to  cut  the  area  into  strips  parallel  to  the 
jc-axis,  the  area  is 


A 


=    11  dy  dx, 


the  limits  in  the  first  integration  being  the  values  of  x  at  the 
ends  of  a  variable  strip;  those  in  the  second  integration,  the 
values  of  y  giving  the  limiting  strips. 

Example.     Find  the  area  bounded  by  the  parabola  y2  = 
4  ax  +  4  a2  and  the  straight  line  y  =  2  a  —  x  (Fig.  466). 


Fig.  46a. 


Fig.  466. 


Solving  simultaneously,  we  find  that  the  parabola  and  the 
line  intersect  at  A  (0,  2  a)  and  B  (8  a,  —  6  a).  Draw  the 
strips  parallel  to  the  z-axis.     The  area  is 


/2a     ria-y  P2a   , '■ 

I         dydx  =    I      [2  a  —  y 


4a 


The  limits  in  the  first  integration  are  the  values  of  x  at  R 
and  S,  the  ends  of  the  variable  strip.  The  limits  in  the 
second  integration  are  the  values  of  y  at  B  and  A,  corre- 
sponding to  the  outside  strips. 


Art.  47 


Volume  by  Double  Integration 


99 


47.  Volume  by  Double  Integration.  —  To  find  the 
volume  under  a  surface  z  =  f  (x,  y)  and  over  a  given  region 
in  the  xy-plsme. 

The  volume  of  the  prism  PQ  standing  on  the  base  Ax  A?/ 
|    (Fig.  47a)  is 

z  Arc  A?/. 

The  volume  of  the  plate  RT  is  then 


Xs  rF  (x) 

z  Ax  Ay  =  Ax   /         z  dy, 

R  J  fix) 


f  (x),  F  (x)  being  the  values  of  y  at  R,  S.     The  entire  volume 
is  the  limit  of  the  sum  of  such  plates 

Xb  PF(x)  fb     pF(x) 

Ax   /         zdy  —    I      I         zdxdy, 

ax=v  a  J  fix)  J  a     J f  (x) 

a,  b  being  the  values  of  x  corresponding  to  the  outside  plates. 
Example.     Find    the    volume    bounded    by    the    surface 
az  =  a2  —  x2  —  4  y2  and  the  x?/-plane. 


Fig.  47a. 


Fig.  476. 


Fig.  476  shows  one-fourth  of  the  required  volume. 
y  =  0.     At  S,  z  =  0  and  so 


At  R, 


y 


=  \Vtf  = 


x~ 


100 


Double  Integration 


Chap.  8 


The  limiting  values  of  x  at  0  and  A  are  0  and  a.     Therefore 

-  (a2  —  x2  —  4  ?/2)  cfa;  dy 


v  =  \\     I  zdxdy  =  4  /      / 

3  a  Jo 


a 


2-x2pdz  = 


48.   The   Double   Integral   as   the   Limit   of   a   Double 

Summation.  —  Divide  a  plane  area  by  lines  parallel  to  the 

coordinate  axes  into  rectangles  with  sides  Ax  and  Ay.     Let 

(x,  y)  be  any  point  within  one  of  these  rectangles.     Form 

the  product 

/  0,  y)  Ax  Ay. 

This  product  is  equal  to  the  volume  of  the  prism  standing 
on  the  rectangle  as  base  and  reaching  the  surface  z  =  f  (x,  y) 
at  some  point  over  the  base.     Take  the  sum  of  such  products 


Fig.  48a. 


for  all  the  rectangles  that  lie  entirely  within  the  area, 
represent  this  sum  by  the  notation 


We 


XX  f^  y) Ax  Ay- 


When  Ax  and  Ay  are  taken  smaller  and  smaller,  this  sum 
approaches  as  limit  the  double  integral 


// 


/  (x,  y)  dx  dy, 


Art.  48 


The  Limit  of  a  Double  Summation 


101 


with  the  limits  determined  by  the  given  area;  for  it  approaches 
the  volume  over  the  area  and  that  volume  is  equal  to  the 
double  integral. 

Whenever  then  a  quantity  is  a  limit  of  a  sum  of  the  form 

X%f(x>y)^xAy 


its  value  can  be  found  by  double  integration.  Furthermore, 
in  the  formation  of  this  sum,  infinitesimals  of  higher  order 
than  Ax  A?/  can  be  neglected  without 
changing  the  limit.  For,  if  e  Ax  Ay 
is  such  an  infinitesimal,  the  sum  of 
the  errors  thus  made  is 


2)  2)  €  Ax  Ay. 


Fig.  486. 


When  Ax  and  Ay  approach  zero,  e 
approaches  zero.  The  sum  of  the 
errors  approaches  zero,  since  it  is 
represented  by  a  volume  whose  thick- 
ness approaches  zero. 

Example  1.     An  area  is  bounded 
by  the  parabola  y2  =  4  ax  and  the 
line    x  =  a.      Find    its    moment    of 
inertia  about  the  axis  perpendicular   to   its   plane  at  the 
origin. 

Divide  the  area  into  rectangles  Ax  Ay.  The  distance  of 
any  point  P  (x,  y)  from  the  axis  perpendicular  to  the  plane 
at  0  is  R  =  OP  =  Vx2  +  y2.  If  then  (x,  y)  is  a  point 
within  one  of  the  rectangles,  the  moment  of  inertia  of  that 
rectangle  is 

B?  Ax  Ay  =  (x2  +  y2)  Ax  Ay, 

approximately.  That  the  result  is  approximate  and  not 
exact  is  due  to  the  fact  that  different  points  in  the  rectangle 
differ  slightly  in  distance  from  the  axis.     This  difference  is, 


102 


Double  Integration 


Chap.  8 


however,  infinitesimal  and,  since  R2  is  multiplied  by  Ax  Ay, 
the  resulting  error  is  of  higher  order  than  Ax  Ay.  Hence  in 
the  limit 

0     J_2V7:{X2  +  y2)dXdy   =   W. 


a\ 


Ex.  2.  Find  the  center  of  grav- 
ity of  the  area  bounded  by  the 
parabolas  y2  =  4  x  +  4,  y2  =  —  2  x 
+  4. 

By  symmetry  the  center  of 
gravity  is  seen  to  be  on  the 
z-axis.     Its  abscissa  is 


_  / 


xdA 


x  = 


A 


If  we  wish  to  use  double  inte- 
gration we  have  merely  to  replace 
dA  by  dx  dy  or  dy  dx.  From  the 
figure  it  is  seen  that  the  first 
integration  should  be  with  respect  to  x.     Hence 


Fig.  48c. 


x  = 


X2   ru 


d-v2)  16 

xdy  dx      -r- 

-4) _  _£_ 

~  8 


/      /  dy 


dx 


2 

5" 


EXERCISES 

Find  the  values  of  the  following  double  integrals : 

J'2    dx  dy 
.  ( 

-2 


'•  r  r 


(x  +  y)2 

/»2tt    ra 

2.     |        I         rdedr. 

Jo      Jas'md 

J        xy  dx  dy. 


4.  f  *   f    e-W  rdddr. 

Jo       Jo 

5.  f3  f  (x2  +  s/2)  dy  dx. 

./0    Jy 


ra     r  Vcfi-y2 

6.     (      |  dydx. 

Jq    Jo 


Art.  49  Double  Integration.    Polar  Coordinates  103 

7.  Find  the  area  bounded  by  the  parabola  if  =  2  x  and  the  line 
x  =  y. 

8.  Find  the  area  bounded  by  the  parabola  y2  =  4:  ax,  the  line 
x  +  y  =  3  a,  and  the  .r-axis. 

9.  Find  the  area  enclosed  by  the  ellipse 

(y-x)2-\-x2  =  1. 

'  10.    Find  the  volume  under  the  paraboloid  z  =  4  —  x2  —  y2  and  over 
the  square  bounded  by  the  lines  x  =  ±\,  2/  =  =fc  1  in  the  xy-plane. 

11.  Find  the  volume  bounded  by  the  z#-plane,  the  cylinder  x2  +  y2 
=  1,  and  the  plane  x  +  y  +  z  =  3. 

12.  Find  the  volume  in  the  first  octant  bounded  by  the  cylinder 
(x  —  l)2  +  (y  —  l)2  =  1  and  the  paraboloid  xy  =  z. 

13.  Find  the  moment  of  inertia  of  the  triangle  bounded  by  the 
coordinate  axes  and  the  line  x  +  y  =  1  about  the  line  perpendicular  to 
it<  plane  at  the  origin. 

14.  Find  the  moment  of  inertia  of  a  square  of  side  a  about  the  axis 
perpendicular  to  its  plane  at  one  corner. 

15.  Find  the  moment  of  inertia  of  the  triangle  bounded  by  the  lines 
x  +  y  =  2,  x  =  2,  y  =  2  about  the  x-axis. 

16.  Find  the  moment  of  inertia  of  the  area  bounded  by  the  parab- 
ola if  =  ax  and  the  line  x  =  a  about  the  line  y  =  —  a. 

17.  Find  the  moment  of  inertia  of  the  area  bounded  by  the  hyper- 
bola xy  =  4  and  the  line  x  +  y  =  o  about  the  line  y  =  x. 

18.  Find  the  moment  of  inertia  of  a  cube  about  an  edge. 

19.  A  wedge  is  cut  from  a  cylinder  by  a  plane  passing  through  a 
diameter  of  the  base  and  inclined  45°  to  the  base.  Find  its  moment  of 
inertia  about  the  axis  of  the  cylinder. 

20.  Find  the  center  of  gravity  of  the  triangle  formed  by  the  lines 
x  =  y,  x  +  y  =  ±,  x  —  2  1/  =  4. 

21.  Find  the  center  of  gravity  of  the  area  bounded  by  the  parabola 
y2  =  4  ax  +  4  a2  and  the  line  y  =  2  a  —  x. 

49.  Double  Integration.  Polar  Coordinates.  —  Pass 
through  the  origin  a  series  of  lines  making  with  each  other 
equal  angles  A0.  Construct  a  series  of  circles  with  centers 
at  the  origin  and  radii  differing  by  Ar.  The  lines  and  circles 
divide  the  plane  into  curved  quadrilaterals  (Fig.  49a). 

Let  r,  6  be  the  coordinates  of  P,  r  +  Ar,  6  +  A0  those  of 
Q.  Since  PU  is  the  arc  of  a  circle  of  radius  r  and  subtends 
the  angle  16  at  the  center,   PR  =  r  Id.     Also  RQ  =  Ar. 


104  Double  Integration  Chap.  8 

When  Ar  and  A0  are  very  small  PRQ  will  be  approximately 
a  rectangle  with  area 

PR-RQ  =  r  A0  Ar. 


Fig.  49a. 

It  is  very  easy  to  show  that  the  error  is  an  infinitesimal  of 
higher  order  than  A0  Ar.  (See  Ex.  5,  page  107.)  Hence 
the  sum 

taken  for  all  the  rectangles  within  a  curve,  gives  in  the  limit 
the  area  of  the  curve  in  the  form 

A  =         I rdddr.  (49a) 

The  limits  in  the  first  integration  are  the  values  of  r  at 
the  ends  A,  B  of  the  strip  across  the  area.  The  limits  in  the 
second  integration  are  the  values  of  0  giving  the  outside 
strips. 

If  it  is  more  convenient  the  first  integration  may  be  with 
respect  to  0.     The  area  is  then 


-//■ 


A  =    I    I  rdrdd. 


Art.  49 


Double  Integration.    Polar  Coordinates 


105 


The  first  limits  are  the  values  of  0  at  the  ends  of  a  strip 
between  two  concentric  circles  (Fig.  496).  The  second  limits 
are  the  extreme  values  of  r. 


Fig.  496. 


Fig.  49c. 


The  element  of  area  in  polar  coordinates  is 

dA  =  rdd  dr. 


(49b) 


We  can  use  this  in  place  of  dA  in  finding  moments  of  inertia, 
volumes,  centers  of  gravity,  or  any  other  quantities  expressed 
by  integrals  of  the  form 


f> 


J{r,6)dA. 
Example  1.     Change  the  double  integral 

Jo     Jo 


,V2 


ax  —  x- 


(x2  +  y2)  dx  dy 


to  polar  coordinates. 

The  integral  is  taken  over  the  area  of  the  semicircle 
y  =  V2  ax  —  x2  (Fig.  49c).  In  polar  coordinates  the 
equation  of  this  circle  is  r  =  2  a  cos  6.     The  element  of  area 


106  Double  Integration  Chap.  8 

dx  dy  can  be  replaced  by  r  dd  dr.*    Also  x2  +  y2  =  r2.     Hence 

J ^2  a    r  V2az-x2  P2     f*2  a  cos  d 

/  (x2  +  y2)dxdy  =    ]      /  r2 .  r  dd  dr. 

o    Jo  Jo   Jo 

The  limits  for  r  are  the  ends  of  the  sector  OP.     The  limits 

for  0  give  the  extreme  sectors  6  =  0,  d  =  -• 

Ex.  2.     Find  the  moment  of  in- 
ertia of  the   area   of   the   cardioid 
r  =  a    (1  +  cos  6)    about   the   axis 
.x  perpendicular  to   its   plane  at  the 
origin. 

The  distance  from  any  point  P 
(r,  8)  (Fig.  49d)  to  the  axis  of  rota- 
tion is 

OP  =  r. 


Fig.  49d. 
Hence  the  moment  of  inertia  is 


7  =  2 


n 

Jo  Jo 


a  (1-fcosO) 


r2  •  r  dd  dr  = 


2Jo   k 


35 

l  +  cos0)4d0  =  —  Tra4. 

lo 


Ex.  3.  Find  the  center  of  gravity  of  the  cardioid  in  the 
preceding  problem. 

The  ordinate  of  the  center  of  gravity  is  evidently  zero. 
Its  abscissa  is 

//V    Pa  (l+cos0) 
xdA      2    /     /  r  cos  0  •  r  dd  dr 

Jo  Jo 


X  = 


fdA 


If 


r  dd  dr 


5 

O 


Ex.  4.  Find  the  volume  common  to  a  sphere  of  radius 
2  a  and  a  cylinder  of  radius  a,  the  center  of  the  sphere  being 
on  the  surface  of  the  cylinder. 

*  This  does  not  mean  that 

dx  dy  =  r  dd  dr, 

but  merely  that  the  sum  of  all  the  rectangular  elements  in  the  circle  is 
equal  to  the  sum  of  all  the  polar  elements. 


Art.  49  Double  Integration.    Polar  Coordinates 


107 


Fig.  49e  shows  one-fourth  of  the  required  volume.  Take 
a  system  of  polar  coordinates  in  the  xy-pl&ne.  On  the 
element  of  area  r  dd  dr  stands  a  prism  of  height         , 

z  =  V4  a2  —  r2, 
z 


Fig.  49e. 
The  volume  of  the  prism  is  z  •  r  dd  dr  and  the  entire  volume  is 


2  fAn2_~.2\2 


r2)' 


2  a  cos  6 


dd 


^> 


Jr»2     r*2  a  cos  0                                                             /*2  (4  fl2 . 
/  V4  a2-r2-  rdddr  =  4:  I    

=  ^T-       (l-sin3^)^  =  ^a3(37r-4). 

EXERCISES 

Find  the  values  of  the  following  integrals  by  changing  to  polar 
coordinates: 

I  (x2  +  2/2)  <fy  dx.       3.     (      I     e-<*2+^  da;  dy. 

n     •/  0  Ja     Jn 


'0     ^  0 

/*V2az-z2 


,a    /» Vai—Xi 


dxdy.  4.     I      )  Va?-x2-y2dxdy. 

0         •'0  •'O     •'0 

6.    Find  the  area  bounded  by  two  circles  of  radii  a,  a  +  Aa  and  two 

lines  through  the  origin,  making  with  the  initial  line  the  angles  a, 

a.  +  Aa,  respectively.     Show  that  when  Aa  and  Aa  approach  zero,  the 

result  differs  from 

a  Aa  Aa 

by  an  infinitesimal  of  higher  order  than  Aa  Aa. 


108  Double  Integration  Chap.  8 

6.  The  central  angle  of  a  circular  sector  is  2  a.  Find  the  moment 
of  inertia  of  its  area  about  the  bisector  of  the  angle. 

7.  An  area  is  bounded  by  the  circle  r  =  a  V2  and  the  straight 

line  r  =  a  sec  f  d  —  ^  ) .     Find  its  moment  of  inertia  about  the  axis  per- 
pendicular to  its  plane  at  the  origin. 

8.  Find  the  center  of  gravity  of  the  area  in  Ex.  6. 

9.  The  center  of  a  circle  of  radius  2  a  lies  on  a  circle  of  radius  a. 
Find  the  moment  of  inertia  of  the  area  between  them  about  the 
common  tangent. 

10.  Find  the  moment  of  inertia  of  the  area  of  the  lemniscate  r2  = 
2  a2  cos  2  6  about  the  axis  perpendicular  to  its  plane  at  the  origin. 

11.  Find  the  moment  of  inertia  of  the  area  of  the  circle  r  =  2  a 

outside  the  parabola,  r  =  a  sec2  -  about  the  axis  perpendicular  to  its 
plane  at  the  origin. 

12.  Find  the  moment  of  inertia  about  the  y-axis  of  the  area  within 
the  circle  (x  -  a)2  +  (y  -  a)2  =  2  a2. 

13.  The  density  of  a  square  lamina  is  proportional  to  the  distance 
from  one  corner.  Find  its  moment  of  inertia  about  an  edge  passing 
through  that  corner. 

14.  Find  the  moment  of  inertia  of  a  cylinder  about  a  generator. 

15.  Find  the  moment  of  inertia  of  a  cone  about  its  axis. 

16.  Find  the  volume  under  the  spherical  surface  x2  +  y2  +  z2  =  a2 
and  over  the  lemniscate  r2  =  a2  cos  2  0  in  the  :n/-plane. 

17.  Find  the  volume  bounded  by  the  z?/-plane,  the  paraboloid 
az  =  x2  +  V2  and  the  cylinder  x2  +  y2  =  2  ax. 

18.  Find  the  moment  of  inertia  of  a  sphere  of  density  p  about  a 
diameter. 

19.  Find  the  volume  generated  by  revolving  one  loop  of  the  curve 
r  =  a  cos  2  6  about  the  initial  line. 

50.  Area  of  a  Surface.  —  Let  an  area  A  in  one  plane  be 
projected  upon  another  plane.     The  area  of  the  projection  is 

A'  =  A  cos  <f>, 

when  ^  is  the  angle  between  the  planes. 

To  show  this  divide  A  into  rectangles  by  two  sets  of  lines 
respectively  parallel  and  perpendicular  to  the  intersection 
MN  of  the  two  planes.     Let  a  and  b  be  the  sides  of  one  of 


Art.  60 


Area  of  a  Surface 


109 


these  rectangles,  a  being  parallel  to  MN.    The  projection  of 
this  rectangle  will  be  a  rectangle  with  sides 

a'  =  a,        b'  —  b  cos  0, 
and  area 

a'b'  =  ab  cos  4>. 

The  sum  of  the  projections  of  all  the  rectangles  is 

£.a'bf  =  ^ab  cos  <£. 

s 

As    the  rectangles    are    taken    smaller    and    smaller    this 

approaches  as  limit 

A'  =  A  cos  <f>, 

which  was  to  be  proved. 


Fig.  50a. 

To  find  the  area  of  a  curved  surface,  resolve  it  into  elements 
whose  projections  on  a  coordinate  plane  are  equal  to  the 
differential  of  area  dA  in  that  plane.  The  element  of  surface 
can  be  considered  as  lying  approximately  in  a  tangent  plane. 
Its  area  is,  therefore,  approximately 

dA 

COS0 

where  0  is  the  angle  between  the  tangent  plane  and  the 

coordinate  plane  on  which  the  area  is  projected.     The  area 

of  the  surface  is  the  limit 

r  dA 

J  cos  <f> 


s  = 


110 


Double  Integration 


Chap.  8 


The  angle  between  two  planes  is  equal  to  that  between 
the  perpendiculars  to  the  planes.     Therefore  <j>  is  equal  to 


Fig.  506. 

the  angle  between  the  normal  to  the  surface  and  the  co- 
ordinate axis  perpendicular  to  the  plane  on  which  we  project. 
If  the  equation  of  the  surface  is 

F  (x,  y,  z)  =  0, 

the  cosine  of  the  angle  between  its  normal  and  the  z-axis  is 
(Differential  Calculus,  Art.  101) 

dF 

dz 


cos  7  = 


s/( 


dF\2  ,(dFV 
dx 


+&)+.<£! 


The  cosines  of  the  angles  between  the  normal  and  the  z-axis 

In 


dF  dF  dF 

or  y-Sixis  are  obtained  by  replacing  —  by  —   or 


dz    "J    dx  dy  ' 

finding  areas  the  algebraic  sign  is  assumed  to  be  positive. 

-Example  1.     Find  the  area  of  the  sphere  x2  +  y2  +  z2  =  a2 
within  the  cylinder  x2  +  y2  =  ax. 

Project  on  the  xy-plsaie.     The  angle  4>  is  then  the  angle  y 
between  the  normal  to  the  sphere  and  the  2-axis.     Its  cosine  is 

z z 

Vx2  +  y2  +  z2"  a 


cos  7  = 


Area  of  a  Surface  111 

Using  polar  coordinates  in  the  a^-plane, 

z  =  Vd2  -  x2  -  y2  =  Va2  -  r2. 
Hence  the  area  of  the  surface  is 

s=fj!±=4f  f—ardec,r    =2a2(j_2)- 
J  cos  7         Jo    Jo  Vd2  —  r2 

Ex.   2.     Find  the  area  of  the  surface  of  the   cone  y2  + 
z2  =  x2  in  the  first  octant  bouncled  by  the  plane  y  +  z  =  a. 
Project  on  the  yz-plane.     Then  <f>  =  a  and 

x  x  I 

COS  (X    -—    ~~ — =3^^^=^=^^^   — —    ■     :=   — — —  • 

VV2  +  if  +  z2      V2x2      V2 
The  area  on  the  cone  is  therefore 

a2V2 


/        V2dydz  = 

o    Jo 


EXERCISES 

1.  Find  the  area  of  the  triangle  cut  from  the  plane 

x  +  2y  +  Zz  =  6 

by  the  coordinate  planes. 

2.  Find  the  area  of  the  surface  of  the  cylinder  x-  +  y2  =  a2  between 
the  planes  z  =  0,  z  =  mx. 

3.  Find  the  area  of  the  surface  of  the  cone  x2  +  y-  =  z2  cut  out  by 
the  cylinder  x2  +  y2  =  2  arc. 

4.  Find  the  area  of  the  plane  x  -\-y  -\-z  =  2a  in  the  first  octant 
bounded  by  the  cylinder  x2  -\-  y2  =  a2. 

5.  Find  the  area  of  the  surface  z2  =  2  xy  above  the  xy-plane  bounded 
by  the  planes  y  =  1,   z  =  2. 

6.  Find  the  area  of  the  surface  of  the  cylinder  x2  -\-  y2  =  2  ax 
between  the  zy-plane  and  the  cone  x2  +  y2  =  z2. 

7.  Find  the  area  of  the  surface  of  the  paraboloid  y2  +  z2  =  2  ax, 
intercepted  by  the  parabolic  cylinder  y2  =  ax  and  the  plane  x  =  a. 

8.  Find  the  area  intercepted  on  the  cylinder  in  Ex.  4. 

9.  A  square  hole  of  side  a  is  cut  through  a  sphere  of  radius  a.  If 
the  axis  of  the  hole  is  a  diameter  of  the  sphere,  find  the  area  of  the 
surface  cut  out. 


CHAPTER  IX 


TRIPLE   INTEGRATION 


51.  Triple  Integrals.  —  The  notation 

I    I    If (x> Vi z) dx dv dz 

*Jxi    *Jyi     *J  Z\ 


is  used  to  represent  the  result  of  integrating  first  with  respect 
to  z  (leaving  x  and  y  constant)  between  the  limits  2i  and  ;%, 
then  with  respect  to  y  (leaving  x  constant)  between  the 
limits  yi  and  y2,  and  finally  with  respect  to  x  between  the 
limits  Xi  and  x%. 


Fig.  52a. 

52.  Rectangular    Coordinates.  —  Divide    a    solid    into 

rectangular  parallelepipeds  of  volume  Ax  Ay  Az  by  planes 
parallel  to  the  coordinate  planes.     To  find  the  volume  of 

112 


Art.  62  Rectangular  Coordinates  113 

the  solid,  first  take  the  sum  of  the  parallelepipeds  in  a  vertical 
column  PQ.     The  result  is 


V  Ax  Ay  Az  =  Ax  Ay   I     dz, 


Zi  and  Z2  being  the  values  of  z  at  the  ends  of  the  column. 
Then  sum  these  columns  along  a  base  MN  and  so  obtain  the 
volume  of  the  plate  MNR.     The  result  is 

lim  YAxAy  I     dz  =  Ax  J       J     dy  dz, 

7/1  and  y<t  being  the  limiting  values  of  y  in  the  plate.  Finally, 
take  the  sum  of  these  plates.     The  result  is  the  triple  integral 

v  =  lim    x  -^x   I       I     dy  dz  =    J       J       I     dxdy  dz, 

Xi,  x%  being  the  limiting  values  of  x. 

It  may  be  more  convenient  to  begin  by  integrating  with 
respect  to  x  or  1/.  In  any  case  the  limits  can  be  obtained 
from  the  consideration  that  the  first  integration  is  a  summa- 
tion of  parallelepipeds  to  form  a  prism,  the  second  a  summa- 
tion of  prisms  to  form  a  plate,  and  the  third  integration  a 
summation  of  plates. 

Let  (x,  y,  z)  be  any  point  of  the  parallelepiped  Ax  Ay  Az. 
Multiply  Ax  Ay  Az  by  /  (x,  y,  z)  and  form  the  sum 

2ZXf(x>y>z)AxAyAz 

taken  for  all  parallelepipeds  in  the  solid.  When  Ax,  Ay,  and 
Az  approach  zero,  this  sum  approaches  the  triple  integral 


/// 


/  (x,  y,  z)  dx  dy  dz 

as  limit.  It  can  be  shown  that  terms  of  higher  order  than. 
Ax  Ay  Az  can  be  neglected  in  the  sum  without  changing 
the  limit. 


114 


Triple  Integration 


Chap.  9 

The  differential  of  volume  in  rectangular  coordinates  is 

dv  =  dx  dy  dz. 

This  can  be  used  in  the  formulas  for  moment  of  inertia,  center 
of  gravity,  etc.,  those  quantities  being  then  determined  by 
triple  integration. 

Example  1.     Find  the  volume  of  the  ellipsoid 


x2  ,  y2  .  z2 


a' 


+  62  +  c2       L 


Fig.  52a  shows  one-eighth  of  the  required  volume.    Therefore 

v  =  8  /     /     I  dxdy  dz. 

The  limits  in  the  first  integration  are  the  values  z  =  0  at  P 


an 


d  z  =  c  y  1 2  — 


y 

r^  at  Q.     The  limits  in  the  second 

integration  are  the  values  of  y  at  M  and  N.     At  M,  y  =  0 

/        x2 
and  at  N,  z  =  0,  whence  y  =  b  y  1 ^ .     Finally,  the  limits 

for  x  are  0  and  a.     Therefore 


v  =  8  I      I  J  dx  dy  dz  =  f  xa&c. 

t/  0    */0  c/0 


Fig.  526. 


Ex.  2.     Find  the  center  of  gravity  of  the  solid  bounded 
by  the  paraboloid  y2  -\-  2  z2  =  4  x  and  the  plane  x  =  2. 


Art.  52 


Rectangular  Coordinates 


115 


By  symmetry  y  and  z  are  zero.     The  ^-coordinate  is 
xdv       4   /      /  /  xdzdydx       . 

_  Jo    Jo Jjiy2+2z*) 4 

Idt)  4    I     I     I  dzdy  dx 

The  limits  for  £  are  the  values  z  =  i  (?/2  +  2  z2)  at  P  and 

x  -  2  at  Q.     At  S,  x  =  2  and  y  =  V4  x  -  2  z2  =  V8  -  2  z2. 
The    limits   for    y    are,    therefore,   2/  =  0    at   R    and   ?/  = 

V8  —  2  z2  at  S.     The  limits  for  z  are  z  =  0  at  A  and  z  =  2 
at  B. 

Ex.  3.     Find  the  moment  of  inertia  of  a  cube  about  an 
edge. 

z 


0) 


Fig.  52c. 

Place  the  cube  as  shown  in  Fig.  52c  and  determine  its 
moment  of  inertia  about  the  z-axis.  The  distance  of  any 
point  (x,  y,  z)  from  the  z-axis  is 

R  =  Vx'2  +  y2. 

Hence  the  moment  of  inertia  is 

J^o      Pa      Pa 
I      I     (x2  +  y2)  dx  dydz  =  \  a5, 
o    Jo    Jo 

where  a  is  the  edge  of  the  cube. 


116 


Triple  Integration 


ChaD.  9 


EXERCISES 

1.  Find  by  triple  integration  the  volume  of  the  pyramid  determined 
by  the  coordinate  planes  and  the  plane  x  +  y  -f-  z  =  1. 

2.  Find  the  moment  of  inertia  of  the  pyramid  in  Ex.  1  about  the 
x-axis. 

3.  A  wedge  is  cut  from  a  cylinder  of  radius  a  by  a  plane  passing 
through  a  diameter  of  the  base  and  inclined  45°  to  the  base.  Find  its 
center  of  gravity. 

ifi       z2         x 

4.  Find  the  volume  bounded  by  the  paraboloid  j^  +  —  =  2  -  and 

the  plane  x  =  a. 

5.  Express  the  volume  of  the  cone 

(z  -  l)2  =  X2  +  y2 

in  the  first  octant  as  a  triple  integral  in  6  ways  by  integrating  with  dx, 
dy,  dz,  arranged  in  all  possible  orders. 

6.  Find  the  volume  bounded  by  the  surfaces  y2  =  4  a2  —  3  ax,  y2  = 
ax,  z  =  zkh. 

7.  Find  the  volume  bounded  by  the  cylinder  z2  =  1  —  x  —  y  and 
the  coordinate  planes. 

53.  Cylindrical  Coordinates.  —  Let  M  be  the  projection 
of  P  on  the  x?/-plane.     Let  r,  6  be  the  polar  coordinates  of  M 

in  the  xy-plane.  The  cylindrical 
coordinates  of  P  are  r,  6,  z. 

From  Fig.  53a  it  is  evident  that 

x  =  r  cos  0,         y  =  r  sin  6. 

By  using  these  equations  we  can 
change  any  rectangular  into  a 
cylindrical  equation. 

The  element  of  volume  in  cy- 
lindrical coordinates  is  the  volume 
PQ,  Fig.  536,  bounded  by  two 
cylindrical  surfaces  of  radii  r, 
r  +  Ar,  two  horizontal  planes  z,  z  +  A  2,  and  two  planes 
through  the  2-axis  making  angles  6,  6  +  A#  with  OX.  The 
base  of  PQ  is  equal  to  the  polar  element  MN  in  the  xy- 
plane.     Its  altitude  PR  is  Az.     Hence 

dv  =  rdd  dr  dz.  (53) 


Fig.  53a. 


Art.  63 


Cylindrical  Coordinates 


117 


This  value  of  dv  can  be  used  in  the  formulas  for  volume, 
center  of  gravity,  moment  of  inertia,  etc.     In  problems  con- 


Fig.  536. 


Fig.  53c. 

ncctcd  with  cylinders,  cones,  and  spheres,  the  resulting 
integrations  are  usually  much  easier  in  cylindrical  than  in 
rectangular  coordinates. 


118 


Triple  Integration 


Chap.  9 


Example  1.  Find  the  moment  of  inertia  of  a  cylinder 
about  a  diameter  of  its  base. 

Let  the  moment  of  inertia  be  taken  about  the  rc-axis, 
Fig.  53c.  The  square  of  the  distance  from  the  element  PQ 
to  the  a>axis  is  _> 

R2  =  y2  -\-  z2  =  r2  sin2  6  +  z2. 

The  moment  of  inertia  is  therefore 

R2dv  =    /        /      /     (r2  sin2  6  +  z2)  r  dd  dz  dr 

*Jo      Jo    Jo 


ira2h 
~12~ 


(3a2  +  4/i2). 


The  first  integration  is  a  summation  for  elements  in  the 
wedge  RS,  the  second  a  summation  for  wedges  in  the  slice 
OMN,  the  third  a  summation  for  all  such  slices. 

Ex.  2.     Find  the  volume  bounded  by  the  xy-plane,  the 
cylinder  x2  +  y2  =  ax,  and  the  sphere  x2  +  y2  +  z2  =  a2. 


Fig.  53d. 


In  cylindrical  coordinates,  the  equations  of  the  cylinder 
and  sphere  are  r  =  a  cos  6  and  r2  +  z2  =  a2.  The  volume 
required  is  therefore 


v  =  2 


J ^2     pa  cos  d     f* 
0      t/0  t^O 


0     f*  Vat—r2 


rdddrdz  =  ia3(3ir  -4). 


Art.  54 


Spherical  Coordinates 


119 


54.  Spherical  Coordinates.  —  The  spherical  coordinates 
of  the  point  P  (Fig.  54a)  are  r  =  OP  and  the  two  angles  6  and 
<f>.  From  the  diagram  it  is 
easily  seen  that 

x  =  r  sin  </>  cos  6, 
y  =  r  sin  <f>  sin  6, 
z  =  r  cos  <j>. 

The  locus  r  =  const,  is  a 
sphere  with  center  at  0;  8  = 
const,  is  the  plane  through  OZ 
making  the  angle  6  with  OX; 
0  =  const,  is  the  cone  gener- 
ated   by   lines    through   0   making  the  angle  0  with  OZ. 

The  element  of  volume  is  the  volume  PQRS  bounded 


Fig.  54a. 


Fig.  546. 

by  the  spheres  r,  r  +  Ar,  the  planes  6,  6  +  Ad,  and  the  cones 
4>,   4>  +  A(p.     When   Ar,   A<f>,  and  A0  are  very  small  this  is 


120  Triple  Integration  Chap.  9 

approximately  a  rectangular  parallelepiped.     Since  OP  =  r 
and  POR  =  A</>, 

PR  =  r  A</>. 

Also  OM  =  OP  sin  4>  and  the  arc  PS  is  approximately  equal 
to  its  projection  MN,  whence 

PS  =  MN  =  r  sin  0  A0, 

approximately.     Consequently 

Av  =  PR-PS-PQ  =  r2  sin  <f>  A0  .  A0  .  Ar, 

approximately.     When   the   increments   are   taken   smaller 

and  smaller,  the  result  becomes  more  and  more  accurate. 

Therefore 

dv  =  r2  sin  0  dd  d(f>  dr.  (54) 

Spherical  coordinates  work  best  in  problems  connected 
with  spheres.  They  are  also  very  useful  in  problems  where 
the  distance  from  a  fixed  point  plays  an  important  role. 


Fig.  54c. 

Example.  If  the  density  of  a  solid  hemisphere  varies  as 
the  distance  from  the  center,  find  its  center  of  gravity. 

Take  the  center  of  the  sphere  as  origin  and  let  the  2-axis 
be  perpendicular  to  the  plane  face  of  the  hemisphere.  By 
symmetry  it  is  evident  that  x  and  y  are  zero.     The  density 


Art.  55  Attraction  121 

is  p  =  kr,  where  k  is  constant.     Also  z  =  r  cos  </>.     Hence 

/  zdm        /  krzdv 


fdm        J 


kr  dv 


Jr»27r      P2       f*a 
o      Uq    *7o 


r4  cos  4>  sin  <f>  dd  d<f>  dr      9 


J^2tt      /*2      /»o 
/      /     r3  sin  0  c?0  d<£  dr 
o      t/o    «7o 


=  -=  a. 
o 


EXERCISES 

1.  Find  the  volume  bounded  by  the  sphere  x2  +  V2  +  22  =  4  and 
the  paraboloid  re2  +  2/2  =  3  z. 

2.  A  right  cone  is  scooped  out  of  a  right  cylinder  of  the  same  height 
and  base.  Find  the  distance  of  the  center  of  gravity  of  the  remainder 
from  the  vertex. 

3.  Find  the  volume  bounded  by  the  surface  z  =  e-W+v2)  and  the 
zy-plane. 

4.  Find  the  moment  of  inertia  of  a  cone  about  a  diameter  of  its  base. 

5.  Find  the  volume  of  the  cylinder  x2  +  if  =  2  ax  intercepted 
between  the  paraboloid  x2  +  y2  =  2  az  and  the  :n/-plane. 

6.  Find  the  center  of  gravity  of  the  volume  common  to  a  sphere  of 
radius  a  and  a  cone  of  vertical  angle  2  a,  the  vertex  of  the  cone  being  at 
the  center  of  the  sphere. 

7.  Find  the  center  of  gravity  of  the  volume  bounded  by  a  spherical 
surface  of  radius  a  and  two  planes  passing  through  its  center  and  in- 
cluding an  angle  of  60°. 

8.  The  vertex  of  a  cone  of  vertical  angle  -x  is  on  the  surface  of  a 

sphere  of  radius  a.  If  the  axis  of  the  cone  is  a  diameter  of  the  sphere, 
find  the  moment  of  inertia  of  the  volume  common  to  the  cone  and 
sphere  about  this  axis. 

55.  Attraction.  —  Two  particles  of  masses  mh  m^  sepa- 
rated by  a  distance  r,  attract  each  other  with  a  force 

kmirrh 


122  Triple  Integration  Chap.  9 

where  &  is  a  constant  depending  on  the  units  of  mass,  dis- 
tance, and  force  used.  A  similar  law  expresses  the  attraction 
or  repulsion  between  electric  charges. 

To  find  the  attraction  due  to  a  continuous  mass,  resolve 
it  into  elements.  Each  of  these  attracts  with  a  force  given 
by  the  above  law.  Since  the 
forces  do  not  all  act  in  the 
same  direction  we  cannot  ob- 
tain the  total  attraction  by 
merely  adding  the  magnitudes 
of  the  forces  due  to  the  sev-  FlG'  55a' 

eral  elements.  The  forces  must  be  added  geometrically. 
For  this  purpose  we  calculate  the  sum  of  the  components 
along  each  coordinate  axis.  The  force  having  these  sums 
as  components  is  the  resultant  attraction. 

If  dm  is  the  mass  of  an  element  at  P,  r  its  distance  from  0, 
and  6  the  angle  between  OX  and  OP,  the  attraction  between 
this  element  and  a  unit  particle  at  0  is 

1  •  dm      k  dm 


k 


/Y>2,  fy*2. 


This  force  acts  along  OP.     Its  component  along  OX  is 

cos  6  •  k  dm 


>'— 


The  component  along  OX  of  the  total  attraction  is  then 

0  dm 


X 


/k  COS! 
r2 


The  calculation  of  this  integral  may  involve  single,  double, 
or  triple  integration,  depending  on  the  form  of  the  attracting 
mass. 

Example  1.  Find  the  attraction  of  a  uniform  wire  of 
length  2  I,  and  mass  M  on  a  particle  of  unit  mass  at  distance 
c  along  the  perpendicular  at  the  center  of  the  wire. 

Take  the  origin  at  the  unit  particle  and  the  x-axis  perpen- 
dicular to  the  wire.     Since  particles  below  OX  attract  down- 


Art.  55 


Attraction 


123 


ward  just  as  much  as  those  above  OX  attract  upward,  the 
vertical  component  of  the  total  attraction  is  zero.  The 
component  along  OX  is,  therefore,  the  total  attraction. 
The  mass  of  the  length  dy  of  the  wire  is 

Mdy 


Hence 


X  = 


21 

kM    fcos  6  dy 
21 


f- 


For  simplicity  of  integration  it  is  better  to  use  0  as  variable. 
Then  y  =  c  tan  6,  dy  =  c  sec2  6  dd,  and 

A 


X  = 


kM  r<* 

21  J-a 


Fig.  556. 


cos  6  •  c  sec2  0  dd      kM 


21  J-a  &  sec2  6               cl 

where  a  is  the  angle  XOA.  In  terms  of  I  this  is 

v_  kM 
X  — 


sin  a, 


c  Vc2  +  I2 

Ex.  2.  Find  the  attraction  of  a  homogeneous  cylinder  of 
mass  M  upon  a  particle  of  unit  mass  on  the  axis  at  distance  c 
from  the  end  of  the  cylinder. 

By  symmetry  it  is  clear  that  the  total  attraction  will  act 
along  the  axis  of  the  cylinder.  Take  the  origin  at  the  attract- 
ing particle  and  let  the  ?/-axis  be  the  axis  of  the  cylinder. 


124 


Triple  Integration 


Chap.  9 


Divide  the  cylinder  into  rings  generated  by  rotating  the 
elements  dx  dy  about  the  y-axis.  The  volume  of  such  a 
ring  is 

2  irx  dx  dy 

and  its  mass  is 


dm  = 


M 

iraVi 


2-Kxdxdy  =  —^-xdxdy. 
a2h 


Since  all  points  of  this  ring  are  at  the  same  distance  from  0 
and  the  joining  lines  make  the  same  angle  6  with  OY,  the 
vertical  component  of  attraction  is 


,   rcos  6  dm  _  ,    Cydm  _  2Mk    f  c+*   A 
J        r2  J     r3  a2h   Jc      Jo 


xy  dy  dx 


2Mk 

a2h 


[h  +  v^+7  -  Va2  +  (c  +  /i)2]. 


Art.  65  Attraction  125 

EXERCISES 

1.  Find  the  attraction  of  a  uniform  wire  of  mass  M  and  length  I  on 
a  particle  of  unit  mass  situated  in  the  line  of  the  wire  at  distance  c  from 
its  end. 

2.  Find  the  attraction  of  a  wire  of  mass  M  bent  in  the  form  of  a 
semicircle  of  radius  a  on  a  unit  particle  at  its  center. 

3.  Find  the  attraction  of  a  flat  disk  of  mass  M  and  radius  a  on  a 
unit  particle  at  the  distance  c  in  the  perpendicular  at  the  center  of  the 
disk. 

4.  Find  the  attraction  of  a  homogeneous  cone  upon  a  unit  particle 
situated  at  its  vertex. 

5.  Show  that,  if  a  sphere  is  concentrated  at  its  center,  its  attraction 
upon  an  outside  particle  will  not  be  changed. 

6.  Find  the  attraction  of  a  homogeneous  cube  upon  a  particle  at 
one  corner. 


CHAPTER  X 
DIFFERENTIAL  EQUATIONS 

56.  Definitions.  —  A  differential  equation  is  an  equation 
containing  differentials  or  derivatives.     Thus 

(x2  +  V2)  dx  +  2  xy  dy  =  0, 
xd?y_dy  =  g 
dx2      dx 

are  differential  equations. 

A  solution  of  a  differential  equation  is  an  equation  connect- 
ing the  variables  such  that  the  derivatives  or  differentials 
calculated  from  that  equation  satisfy  the  differential  equa- 
tion. Thus  y  =  x2  —  2  x  is  a  solution  of  the  second  equation 
above;  for  when  x2  —  2  x  is  substituted  for  y  the  equation  is 
satisfied. 

A  differential  equation  containing  only  a  single  independent 
variable,  and  so  containing  only  total  derivatives,  is  called 
an  ordinary  differential  equation.  An  equation  containing 
partial  derivatives  is  called  a  partial  differential  equation. 
We  shall  consider  only  ordinary  differential  equations  in  this 
book. 

The  order  of  a  differential  equation  is  the  order  of  the 
highest  derivative  occurring  in  it. 

57.  Illustrations  of  Differential  Equations.  —  Whenever 
an  equation  connecting  derivatives  or  differentials  is  known, 
the  equation  connecting  the  variables  can  be  determined  by 
solving  the  differential  equation.  A  number  of  simple  cases 
were  treated  in  Chapter  I. 

The  fundamental  problem  of  integral  calculus  is  to  find 
the  function 


V  =  ^ 


j  fix)  dx, 


Art.  57 


Illustrations  of  Differential  Equations 


127 


when  /  (x)  is  given.     This  is  equivalent  to  solving  the  differ- 
ential equation 

dy  =  f  (x)  dx. 

Often  the  slope  of  a  curve  is  known  as  a  function  of  x  and  y, 

dy 


dx 


=  f(x,y)- 


The  equation  of  the  curve  can  be  found  by  solving  the 
differential  equation. 

In  mechanical  problems  the  velocity  or  acceleration  of  a 
particle  may  be  known  in  terms  of  the  distance  s  the  particle 
has  moved  and  the  time  t, 


ds 
dt=V> 


dh 

dt2 


=  a. 


The  position  s  can  be  determined  as  a  function  of  the  time  by 
solving  the  differential  equation. 

In  physical  or  chemical  problems  the  rates  of  change  of 
the  variables  may  be  known  as  functions  of  the  variables 
and  the  time.  The  values  of  those  variables  at  any  time  can 
be  found  by  solving  the  differential  equations. 

Example.  Find  the  curve  in  which  the  cable  of  a  suspension 
bridge  hangs. 


Fk;.  57 


Let  the  bridge  be  the  z-axis  and  let  the  ?/-axis  pass  through 
the  center  of  the  cable.     The  portion  of  the  cable  AP  is  in 


128  Differential  Equations  Chap.  10 

equilibrium  under  three  forces,  a  horizontal  tension  H  at  A, 
a  tension  PT  in  the  direction  of  the  cable  at  P,  and  the 
weight  of  the  portion  of  the  bridge  between  A  and  P.  The 
weight  of  the  cable,  being  very  small  in  comparison  with  that 
of  the  bridge,  is  neglected. 

The  weight  of  the  part  of  the  bridge  between  A  and  P  is 
proportional  to  x.  Let  it  be  Kx.  Since  the  vertical  com- 
ponents of  force  must  be  in  equilibrium 

TsiiKf)  =  Kx. 

Similarly,  from  the  equilibrium  of  horizontal  components, 
we  have 

T  cos  0  =  H. 
Dividing  the  former  equation  by  this,  we  get 


*      ^      K 
tan  (f>  =  yz  x. 

Jti 


dti 
But  tan  <f>  =  -j- .     Hence 

dy_K 
dx      HX' 

The  solution  of  this  equation  is 

K     „ 

y  =  2HX+C- 

The  curve  is  therefore  a  parabola. 

58.   Constants  of  Integration.     Particular  and  Genera) 
Solutions.  —  To  solve  the  equation 

we  integrate  once  and  so  obtain  an  equation  with  one  arbi- 
trary constant, 


V  =   f  f  0)  dx  +  c. 


Art.  58  Constants  of  Integration  129 

To  solve  the  equation 

we  integrate  twice.     The  result 

y  =  J    jf(x)  dx2  +  ax  +  c2 

contains  two  arbitrary  constants.     Similarly,  the  integral  of 
the  equation 

dxn     J  v  J 

contains  n  arbitrary  constants. 

These  illustrations  belong  to  a  special  type.  The  rule 
indicated  is,  however,  general.  The  complete,  or  general, 
solution  of  a  differential  equation  of  the  nth  order  in  two  vari- 
ables contains  n  arbitrary  constants.  If  particular  values  are 
assigned  to  any  or  all  of  these  constants,  the  result  is  still  a 
solution.     Such  a  solution  is  called  a  particular  solution. 

In  most  problems  leading  to  differential  equations  the 
result  desired  is  a  particular  solution.  To  find  this  we 
usually  find  the  general  solution  and  then  determine  the 
constants  from  some  extra  information  contained  in  the 
statement  of  the  problem. 

Example  L    Show  that 

x2  +  y2  -  2  ex  =  0 
is  the  general  solution  of  the  differential  equation 

y2  -  x2  -  2  xy  p-  =  0. 

Differentiating  x2  +  y2  —  2  ex  =  0,  we  get 

2x  +  2y^--  2c  =  0, 
dx 

whence 

dy      c  —  x 
dx  y 


130  Differential  Equations  Chap.  10 

Substituting  this  value  in  the  differential  equation,  it  becomes 
y2  —  x2  —  2  xy  -~  =  y2  —  x2  —  2  x  (c  —  x)  =  y2  +  x2  —  2  ex  =  0. 

Hence  x2  +  y2  —  2  ex  =  0  is  a  solution.  Since  it  contains 
one  constant  and  the  differential  equation  is  one  of  the  first 
order,  it  is  the  general  solution. 

Ex.  2.  Find  the  differential  equation  of  which  y  =  Ciex  + 
c2e2x  is  the  general  solution. 

Since  the  given  equation  contains  two  constants,  the 
differential  equation  is  one  of  the  second  order.  We  there- 
fore differentiate  twice  and  so  obtain 

JL  =  Clex  +  2c2e2x, 
ax 

3  - cie* + 4  *"■ 

Eliminating  ch  we  get 

dxj 


dx~v  =  c^'- 


Hence 


or 


d2y      dy 

dx2      dx 


=2(fH 


^_3^  +  2y=0. 
dx-         dx 

This  is  an  equation  of  the  second  order  having  y  =  Ciex  + 
e%e2x  as  solution.     It  is  the  differential  equation  required. 

EXERCISES 

In  each  of  the  following  exercises,  show  that  the  equation  given  is  a 
solution  of  the  differential  equation  and  state  whether  it  is  the  general 
or  a  particular  solution. 

1.    y  =  cex  +  e-*,  -^  =  y. 


'! 


Art.  59  The  First  Order  in  Two  Variables 

2.  x-  -  if  =  c.c, 

3.  y  =  ce*  sin  a:, 


131 


(.r  -+-  >f)  dx  -  2  xy  dy  =  0. 


'4.    ?/  =  d  +  c2  sin  (z  +  c3),  dc^  +  dc         ' 

Find  the  differential  equation  of  which  each  of  the  following  equa- 
tions is  the  general  solution: 

d  v    7.  y  =  Ci  sin  x  -\-  Cz  cos  rr. 

8.  .r2//  =  cx  +  c2  In  x  +  C3X3. 

Q.    y  =  cxcx.  9.  x2  -f  fix?/  +  Ctif  =  0. 


5.    ?/  =  CiX  -\ 

J  x 


59.  Differential  Equations  of  the  First  Order  in  Two 
Variables.  —  By  solving  for  ■—■  an  equation  of  the  first 
order  in  two  variables  x  and  y  can  be  reduced  to  the  form 

To  solve  this  equation  is  equivalent  to  finding  the  curves 
with   slope   equal   to  /  (x,  y) .     The   solution   contains   one 


Fig.  59. 

arbitrary  constant.  There  is  consequently  an  infinite  num- 
ber of  such  curves,  usually  one  through  each  point  of  the 
plane. 

"We  cannot  always  solve  even  this  simple  type  of  equa- 
tion.    In  the  following  articles  some  cases  will  be  discussed 


132  Differential  Equations        '  Chap.  10 

which  frequently  occur  and  for  which  general  methods  of 
solution  are  known. 

60.   Variables   Separable.  —  A   differential   equation  of 
the  form 

Mdx  +  Ndy  =  0 

is  called  separable  if  each  of  the  coefficients  M  and  N  con- 
tains only  one  of  the  variables  or  is  the  product  of  a  function 
of  x  and  a  function  of  y.  By  division  the  x's  and  dx  can  be 
brought  together  in  the  first  term,  the  y's  and  dy  in  the 
second.  The  two  terms  can  then  be  integrated  separately 
and  the  sum  of  the  integrals  equated  to  a  constant. 

Example  1.     (1  +  x2)  dy  —  xy  dx  =  0. 

Dividing  by  (1  +  x2)  y,  this  becomes 

dy        xdx 

y       1  +  x2 ' 
whence 

lny  =  |ln(l+x2)  +  c. 

If  c  =  In  k,  this  is  equivalent  to 

In y  =  lnVl+x2+ln/c  =  Ink  Vl+x2, 


and  so 


y  —  k  V 1  +  x2, 


where  k  is  an  arbitrary  constant. 

Ex.  2.  Find  the  curve  in  which  the  area  bounded  by  the 
curve,  coordinate  axes,  and  a  variable  ordinate  is  proportional 
to  the  arc  forming  part  of  the  boundary 

Let  A  be  the  area  and  s  the  length  of  arc.     Then 

A  =  ks. 

Differentiating  with  respect  to  x, 

dA      ,ds_ 
dx         dx' 
or 

'T^  +  fSf' 


Art.  61  Exact  Differential  Equations  133 


y 


Solving  for  -7-, 


dy       V?y2  —  A;2 

dx~~~k        ? 

whence 

dy        =  dx 

\/y2  —  k2       k 
The  solution  of  this  is 


Inty+Vjf-  K?)  =X 

Therefore 


ln(y  +  V^-^)=|  +  c. 


I 


y  +V|/2  —  k2  =  e*      =  ece*  =  cie*, 
where  Ci  is  a  new  constant.     Transposing  2/  and  squaring, 
we  get 

(i\2  x 

ciek)   —  2c^y  +  f/2. 
Hence,  finally, 

1  72  * 

Cl     jT     ,       ft/"       —  I 

y  =  2e+2J1e 

61.   Exact  Differential  Equations.  —  An  equation 

du  =  0, 

obtained  by  equating  to  zero  the  total  differential  of  a  func- 
tion u  of  x  and  y,  is  called  an  exact  differential  equation. 
The  solution  of  such  an  equation  is 

u  =  c. 

The  condition  that  M  dx  +  N  dy  be  an  exact  differential 
is  (Diff.  Cal.,  Art.  100) 

dM      dN 


dy        dx 
This  equation,  therefore,  expresses  the  condition  that 

M  dx  +  N  dy  =  0 
be  an  exact  differential  equation. 


(61) 


134  Differential  Equations  Chap.  10 

An  exact  equation  can  often  be  solved  by  inspection.  To 
find  u  it  is  merely  necessary  to  obtain  a  function  whose  total 
differential  is  M  dx  +  N  dy. 

If  this  cannot  be  found  by  inspection,  it  can  be  determined 
from  the  fact  that 

du  =  M  dx  +  N  dy 
and  so 

dx 

By  integrating  with  y  constant,  we  therefore  get 

u  =  fMdx+f(y). 

Since  y  is  constant  in  the  integration,  the  constant  of  inte-, 
gration  may  be  a  function  of  y.  This  function  can  be  found 
by  equating  the  total  differential  of  u  to  M  dx  +  N  dy.-' 
Since  df  (y)  gives  terms  containing  y  only,  /  (y)  can  usually 
be  found  by  integrating  the  terms  in  N  dy  that  do  not  contain  x. 
In  exceptional  cases  this  may  not  give  the  correct  result. 
The  answer  should,  therefore,  be  tested  by  differentiation. 

Example  1.     (2  x  —  y)  dx  +  (4  y  —  x)  dy  =  0. 

The  equation  is  equivalent  to 

2  x  dx  +  4  y  dy  —  (y  dx  +  x  dy)  =  d  (x2  +  2  y2  —  xy)  =  0. 

It  is  therefore  exact  and  its  solution  is 

x2  +  2  y2  —  xy  =  c. 


Ex.  2.     (\ny  -2x)dx  +  (--  2y\dy=  0. 


In  this  case 


dM       d  n  o   \-     1 

—  =^-(hiy  -2x)  =-■> 
dy       dyx  y 

dN  =d_(x_2    \=1. 
dx     '  dx  \y  J      y 


Art.  62  Integrating  Factors  135 

These  derivatives  being  equal,  the  equation  is  exact.  Its 
solution  is 

x  In  y—  x2  —  y2  =  c. 

The  part  x  In  y  —  x2  is  obtained  by  integrating  (In  y  —  2  x)  dx 
with  y  constant.     The  term  —  y2  is  the  integral  of  —  2  y  dy, 

which  is  the  only  term  in  ( 2  y )  dy  that  does  not  contain  x. 

62.  Integrating  Factors.  —  If  an  equation  of  the  form 
M  dx  +  Ar  dy  =  0  is  not  exact  it  can  be  made  exact  by 
multiplying  by  a  proper  factor.  Such  a  multiplier  is  called 
an  integrating  factor. 

For  example,  the  equation 

x  dy  —  y  dx  =  0 
is  not  exact.     But  if  it  is  multiplied  by  —  >  it  takes  the  form 


x  dy  —  y  dx 


x2 


-C)-o 


which  is  exact.     It  also  becomes  exact  when  multiplied  by 

—  or  — .     The  functions  -=>  -=>  —  are  all  integrating  factors 
y-      xy  xl   yl   xy 

of  x  dy  —  y  dx  =  0. 

While  an  equation  of  the  form  M  dx  -f  N  dy  =  0  always 
has  integrating  factors,  there  is  no  general  method  of  finding 
them. 

Example  1.     y  (1  +  xy)  dx  —  x  dy  =  0. 

This  equation  can  be  written 

y  dx—  xdy  +  xy2  dx  =  0. 
Dividing  by  y2, 

y  dx  —  x  dy 


y2 


+  x  dx  =  0. 


Both  terms  of  this  equation   are  exact  differentials.     The 
solution  is 

x    i    !    2 

y  +  2X   =C' 


136  Differential  Equations  Chap.  10 

Ex.  2.     (y2  +  2  xy)  dx  +  (2  x2  +  3  zy)  dy  =  0. 
This  is  equivalent  to  » 

y2  dx  +  3  a^/  c/?/  +  2  zy  dx  +  2  z2  cfa/  =  0. 
Multiplying  by  y,  it  becomes 
yz  dx  +  3  #2/2  d?/  +  2  zi/2  dx  -\-  2  x2y  dy  =  d  (xy3  +  z2?/2)  =  0. 

Hence 

z?/3  +  x2y2  =  c. 

63.  Linear  Equations.  —  A  differential  equation  of  the 
form 

i  +  py  =  Q>  (63a) 

where  P  and  Q  are  functions  of  x  or  constants,  is  called 
linear.  The  linear  equation  is  one  of  the  first  degree  in  one 
of  the  variables  (y  in  this  case)  and  its  derivative.  Any 
functions  of  the  other  variable  can  occur. 

If  the  linear  equation  is  written  in  the  form  (63a), 

is  an  integrating  factor;  for  when  multiplied  by  this  factor 
the  equation  becomes 

The  left  side  is  the  derivative  of 

fPdx 

ye 

Hence 

fpdx     r  fpdx 


fefpdxQdx  +  c  (63b) 


is  the  solution. 


Example  1.     -j-  +  -y  =  x3. 
ax      x 


Art.  64  Equations  Reducible  to  Linear  Form  137 

In  this  case 


Hence 


j  Pdx  =    I  -  dx  =  2  In  x  =  In  x2. 


fp  dx  In  i»  « 

e         =  e       =  x2. 


The  integrating  factor  is,  therefore,  x2.     Multiplying  by  x2 
and  changing  to  differentials,  the  equation  becomes 

x2  dy  +  2  xy  dx  =  xb  dx. 

The  integral  is 

x2y  =  I  x6  +  c. 

&c.  2.     (1  +  y2)  dx  -  (xy  +y  +  if)  dy  =  0. 

This  is  an  equation  of  the  first  degree  in  x  and  dx.     Divid- 
ing by  (1  +  y2)  dy,  it  becomes 

dx  y 

P  is  here  a  function  of  y  and 

fpdy  =         1 

Multiplying  by  the  integrating  factor,  the  equation  becomes 

dx  xy  dy  y  dy 

Vl  +  y*  ~  (1  +  */)*  ~~  VT  +  P' 

whence 


Vl+2/2 
and 


=  Vl  +  w2  + 


x 


-  1  +  y2  +  c  VTT?. 


64.   Equations  Reducible  to  Linear  Form.  —  An  equation 
of  the  form 

^  +  Py  =  Or,  (64) 


138  Differential  Equations  Chap.  10 

where  P  and  Q  are  functions  of  x,  can  be  made  linear  by  a 
change  of  variable.     Dividing  by  yn,  it  becomes 


If  we  take 


y-*%  + Py-"+l  =  Q. 


2/i-»  =  u 


as  a  new  variable,  the  equation  takes  the  form 

1      du 


1  —  n  dx 
which  is  linear. 

nil       2  ii 

Example.     Xx  +  xV  =  x3' 


+  Pu  =  Q, 


Division  by  y3  gives 


~dy   .   2     „       1 

J     dx      xu         x3 


u  =  y~2. 


Let 
Then 

dx  "  u     dx ' 


du  _      ,  dy 

=  -2y~3 


whence 

_3dy  1  du 

dx  2  dx 

Substituting  these  values,  we  get 


ldu   ,   2  1 

and  so 


2  dx      x         x3' 


da      4  2 

ri  Of*  i*  /y*3 


This  is  a  linear  equation  with  solution 

1 

3f 


1  4 

^  :    :    Q  ~2  ~T~  CX  > 


or,  since  w  =  y~2, 

I       J-4.      4 

o   '  '    o      o  ~T~  CX  . 

?/2       3r 


Art.  65  Homogeneous  Equations  139 

65.    Homogeneous   Equations.  —  A   function  /  (x,  y)    is 
said  to  be  a  homogeneous  function  of  the  nth  degree  if 

f(tx,ty)  =t"f(x,y). 

Thus   Vx2  +  y2  is  a  homogeneous  function  of  the  first 
degree;  for 

VxH*  +  y-t-  =  t  Vx2  +  y\ 

It  is  easily  seen  that  a  polynomial  whose  terms  are  all  of 
the  ?zth  degree  is  a  homogeneous  function  of  the  nth  degree. 
The  differential  equation 

M  dx  +  N  dy  =  0 

is  called  homogeneous  if  M  and  N  are  homogeneous  functions 
of  the  same  degree.  To  solve  a  homogeneous  equation 
substitute 

y  =  vx. 

The  new  equation  will  be  separable. 

Example  1.     x-, y  =  Vx2  +  V2- 

This  is  a  homogeneous  equation  of  the  first  degree.     Sub- 
stituting y  =  vx,  it  becomes 


(  v  +  x  -j- )  —  vx  =  Vx2  +  vH 


whence 

dv         /- — ; — r 
x  -p  =  Vl  +v2. 

This  is  a  separable  equation  with  solution 

x  =  c(v  +  Vl  +i>2). 

Replacing  y  by  -,  transposing,  squaring,  etc.,  the  equation 

x 

becomes 

x2  —  2  cy  =  c2. 


-140  Differential  Equations  Chap.  10 

dv 
Solving  for  -p,  we  get 

dy      —  x  ±  Vx2  +  y2 
dx  y 

or 

y  dy  +  xdx  =  =L  Vx2  +  y2  dx. 

This  is  a  homogeneous  equation  of  the  first  degree.  It  is 
much  easier,  however,  to  divide  by  Vx2  +  y2  and  integrate 
at  once.     The  result  is 

xdx  +  y  dy  1 

V  x2  +  y2 
whence 

Vx2  +  y2  =  c  ±x 
and 

2/2  =  c2  _j_  2  ex. 

Since  c  may  be  either  positive  or  negative,  the  answer  can  be 
written 

2/2    =   c2  _L.  2  C£. 

66.   Change  of  Variable.  —  We  have  solved  the  homo- 
geneous equation  by  taking  as  new  variable 

„    y 

v  =  -• 

X 

It  may  be  possible  to  reduce  any  equation  to  a  simpler  form 
by  taking  some  function  u  of  x  and  yasa  new  variable  or  by 
taking  two  functions  u  and  v  as  new  variables.  Such  func- 
tions are  often  suggested  by  the  equation.  In  other  cases 
they  may  be  indicated  by  the  problem  in  the  solution  of 
which  the  equation  occurs. 

Example,     (x  —  y)2  -p  =  a2. 


Art.  66  Change  of  Variable 

Let  x—  y  =  u.     Then 

dy      du 
dx      dx 

and  the  differential  equation  becomes 

du 


whence 


u2  —  a2  =  w2 


da; 


The  variables  are  separable.     The  solution  is 

a 


.  a,    u 
x  =  u  +-  In — ■  — 
2      u  -\-  a 


+  c 


.a,    x  —  y  —  a  , 

=  x  —  i/  +  jr  In ^-- — ■  +  c, 

J       2      x  —  y  +a        ' 


or 


a 


y  =  -  In ^— r 

y       2      £  —  2/  +  a 


+  c. 


141 


EXERCISES 

Solve  the  following  differential  equations: 

1.  x3  dy  —  y3  dx  =  0. 

2.  tan  x  sin2  y  dx  -\-  cos2  x  cot  y  dy  =  0. 

3.  (xy2  +  x)  dx  +  (y  -  tfy)  dy  =  o. 

4.    (xy2  +  x)  dx  +  (x2y  —  y)  dy  =  0. 

6.    (3  x2  +  2  xy  --  y2)  dx  +  (x2  -  2  xy  -  3  y2)  dy  =  0. 

12.    (2  xy2  -  y)  dx  +  x  dy  =  0. 


6.  if^y-* 

7.  xdx-\- ydy  =  a(x2  +  y2)dy. 

8.  ,*  +  „..*. 


dx 


9.  -  aw  =  ete. 

dx 

10.  x2  ^  -  2  xy  =  3. 

11.  x2^-2xy  =3y. 


13.  (l-x2)^+2xy  =  (l-x2)2. 

ox 

dy 

14.  tan x-~  —  y  =  a. 

dx 

15.  x^-3y  +  zV  =  0. 

ax 

16.  ^  +  y  *  xy3. 


142  Differential  Equations  Chap.  10 

17.  (x2  -  I)*  dy  +  (x3  +  3  xy  Vx°-  -  l)  dx  =  0. 

18.  x  dx  -\-  (x  -j-  y)  dy  =  0. 

19.  (x2  +  ?/-)  dx  -2xydy  =  0. 

20.  i/  dx  +  (a:  +  y)  dy  =  0. 

21.  (x3  -  3  x2?/)  dx  +  (?/3  -  x3)  dy  =  0. 

22.  2/ey  dx  =  Q/3  +  2  sev)  d?/. 

23.  (x?/^  +  yV  dx  -  x2ey  dy  =  0. 

24.  (x  +  y  -  1)  dx  +  (2  x  +  2  y  -  3)  dy  =  0. 

25.  Sy*$fTy*  =  x. 


28 

30.  The  differential  equation  for  the  charge  q  of  a  condenser  having 
a  capacity  C  connected  in  series  with  a  circuit  of  resistance  R  is 

T>dq       q  _  „ 

Rdt  +  c~E> 

where  E  is  the  electromotive  force.     Find  v  as  a  function  of  t  if  E  is 
constant  and  q  =  0  when  i  =  0. 

31.  The  differential  equation  for  the  current  induced  by  an  electro- 
motive force  E  sin  at  in  a  circuit  having  the  inductance  L  and  resist- 
ance R  is 

di 
L-^  +  Ri  —  E  sin  aZ. 
at 

Solve  for  £  and  determine  the  constants  so  that  i  =  I  when  t  =  0. 

Let  PT  be  the  tangent  and  PN  the  normal  to  a  plane  curve  at 
P  (x,  y)  (Fig.  66a).  Determine  the  curve  or  curves  in  each  of  the 
following  cases: 

32.  The  subtangent  TM  =  3  and  the  curve  passes  through  (2,  2). 

33.  The  subnormal  MN  =  a  and  the  curve  passes  through  (0,  0). 

34.  The  intercept  OT  of  the  tangent  on  the  x-axis  is  one-half  the 
abscissa  OM. 

35.  The  length  PT  of  the  tangent  is  a  constant  a. 

36.  The  length  PN  of  the  normal  is  a  constant  a. 

37.  The  perpendicular  from  M  to  PT  is  a  constant  a. 


Art.  67 


Certain  Equations  of  the  Second  Order 


143 


Using  polar  coordinates  (Fig.  666),  find  the  curve  or  curves  in  each 
of  the  following  cases : 


Y 

1 

->  / 

\     y 

0 

T 

21 

A   A 

Fig.  66a. 


Fig.  666. 


38.    The  curve  passes  through  (1,  0)  and  makes  with  OP  a  constant 
angle  \p 


TV 


39.  The  angles  \p  and  0  are  equal. 

40.  The  distance  from  0  to  the  tangent  is  a  constant  a. 

41.  The  projection  of  OP  on  the  tangent  at  P  is  a  constant  a. 

42.  Find  the  curve  passing  through  the  origin  in  which  the  .area 
bounded  by  the  curve,  rr-axis,  a  fixed,  and  a  variable  ordinate  is  pro- 
portional to  that  ordinate. 

43.  Find  the  curve  in  which  the  length  of  arc  is  proportional  to  the 
angle  between  the  tangents  at  its  end. 

44.  Find  the  curve  in  which  the  length  of  arc  is  proportional  to  the 
difference  of  the  abscissas  at  its  ends. 

45.  Find  the  curve  in  which  the  length  of  any  arc  is  proportional  to 
the  angle  it  subtends  at  a  fixed  point. 

46.  Find  the  curve  in  which  the  length  of  arc  is  proportional  to  the 
difference  of  the  distances  of  its  ends  from  a  fixed  point. 

47.  Oxygen  flows  through  one  tube  into  a  liter  flask  filled  with  air 
while  the  mixture  of  oxygen  and  air  escapes  through  another.  If  the 
action  is  so  slow  that  the  mixture  in  the  flask  may  be  considered  uniform, 
what  percentage  of  oxygen  will  the  flask  contain  after  10  liters  of  gas 
have  passed  through?  (Assume  that  air  contains  21  per  cent  by  volume 
of  oxygen.) 

67.   Certain   Equations   of  the   Second   Order.  —  There 

are  two  forms  of  the  second  order  differential  equation  that 


144  Differential  Equations  Chap.  10 

occur  in  mechanical  problems  so  frequently  that  they  de- 
serve special  attention.     These  are 

»  2 -'(*»• 


(2)  ^y=f(ydy) 

K)  dx2      J  \y'  dx) 


The  peculiarity  of  these  equations  is  that  one  of  the  vari- 
ables (y  in  the  first,  x  in  the  second)  does  not  appear  directly 
in  the  equation.  They  are  both  reduced  to  equations  of  the 
first  order  by  the  substitution 

dy 
This  substitution  reduces  the  first  equation  to  the  form 

This  is  a  first  order  equation  whose  solution  has  the  form 

p  =  F(x,  ci), 


or,  since  p  =  -f-> 


dy 

dy 


dz  =  F(*'Cl>- 


This  is  again  an  equation  of  the  first  order.     Its  solution  is 
the  result  required. 

In  case  of  an  equation  of  the  second  type,  write  the  second 
derivative  in  the  form 

d?y      dp  _  dp    dy         dp 
dx2      dx      dy    dx         dy 

The  differential  equation  then  becomes 

p|=/(2/,p). 


Art.  67  Certain  Equations  of  the  Second  Order  145 

Solve  this  for  p  and  proceed  as  before. 

Example  1.     (1  +  z2)  ^  +  1  +  ^Y  =  0. 


Substituting  p  for  -p ,  we  get 

(1+*2)J  +  1+P2  =  0. 

This  is  a  separable  equation  with  solution 

C\  —  x 

whence 

dy  =  t-, dx. 

U         1  +  CiX 

The  integral  of  this  is 

V=  --  +  ^4^1n  (1  +  c&)  +  c2. 

Ci  Ci2 

By  a  change  of  constants  this  becomes 

y  =  ex  +  (1  +  c2)  In  (c  —  z)  +  c'. 


Substituting 


we  get 


da;  dx2         cfa/ ' 

The  solution  of  this  is 

y2P2  =  y2  +  ci. 

dy 
Replacing  p  by  -^  and  solving  again,  we  get 

y2  +  Ci  =  (a:  +  c2)2. 


146  Differential  Equations  Chap.  10 

Ex.  3.   Under  the  action  of  gravitation  the  acceleration  of 

k 
a  falling  body  is  —2 ,  where  k  is  constant  and  r  the  distance 

from  the  center  of  the  earth.  Find  the  time  required  for  the 
body  to  fall  to  the  earth  from  a  distance  equal  to  that  of  the 
moon. 

Let  rx  be  the  radius  of  the  earth  (about  4000  miles) ,  r2  the 
distance  from  the  center  of  the  earth  to  the  moon  (about 
240,000  miles)  and  g  the  acceleration  of  gravity  at  the  surface 
of  the  earth  (about  32  feet  per  second).  At  the  surface  of 
the  earth  r  =  ri  and 

k 

The  negative  sign  is  used  because  the  acceleration  is  toward 
the  origin  (r  =  0) .  Hence  k  =  —  gri2  and  the  general  value 
of  the  acceleration  is 


The  time  of  falling  is  therefore 


t  =  /     y  9 ttt1       —  dr  =  '.  16  hours. 


7T2 

-  gn2  (r2  -  r) 


This  result  is  obtained  by  using  the  numerical  values  of  n 
and  r2  and  reducing  g  to  miles  per  hour. 


a 

v  dv           gri2       — -~ 
dr             r2 

*~ 

where  v  is 

the  velocity. 

The  solution  of  this 

equation  is 

r 

When  r 

=  r2, 

v  =  0. 

Consequently, 

C=      2gri 
r2 

and 

Art.  68  Constant  Coefficients  147 

68.   Linear  Differential  Equations  with  Constant  Coeffi- 
cients. —  A  differential  equation  of  the  form 

^  +  ai^  +  aa^+  •  •  •  +^=/W'    (68a) 

where  ai,  02,  .  .  .  o«  are  constants,  is  called  a  linear  differen- 
tial equation  with  constant  coefficients.    For  practical  applica- 
tions this  is  the  most  important  type  of  differential  equation. 
In  discussing  these  equations  we  shall  find  it  convenient 

to  represent  the  operation  —  by  D.     Then 

dy       r,  d2y       _ 

£  =  D»>    d  -  D-y> ete- 

Equation  (68a)  can  be  written 

(D»  +  a,D^  +  a2D»-2  +  •  •  •  +  an)y  =f(x).     (68b) 

This  signifies  that  if  the  operation 

Dn  +  ciiD"-1  +  a2Dn~2  +  •  •  •  +  an  (68c) 

is  performed  on  y,  the  result  will  be  /  (x) .  The  operation 
consists  in  differentiating  y,  n  times,  n  —  1  times,  n  —  2 
times,  etc.,  multiplying  the  results  by  1,  ah  a2,  etc.,  and 
adding. 

With  the  differential  equation  is  associated  an  algebraic 
equation 

rn  _|_  aifn-l   _|_  a2rn-2  _|_     .     .     .     _|_  ^   =   Q. 

If  the  roots  of  this  auxiliary  equation  are  ri,  r2,  .  .  .  rn,  the 
polynomial  (68c)  can  be  factored  in  the  form 

•  (D  -  n)  (D  -  r,)   •  •  •  (D  -  rB).  (68d) 

If  we  operate  on  y  with  D  —  a,  we  get 

(D  -  a)  2/  =  ^  -  a?/. 


148  Differential  Equations  Chap.  10 

If  we  operate  on  this  with  D  —  b,  we  get 

(D  -  b)  ■  (D  -  a)  y  =  (D  -  b)  (g  -  ay) 

d?y  dy 

The  same  result  is  obtained  by  operating  on  y  with 

(D  -  a)  (D  -  b)  =  D2  -  (a  +  6)  D  +  ab. 

Similarly,  if  we  operate  in  succession  with  the  factors  of 
(68d),  we  get  the  same  result  that  we  should  get  by  operating 
directly  with  the  product  (68c). 

69.  Equation  with  Right  Hand  Member  Zero.  —  To  solve 
the  equation 

(7>  +  aj)n-1  +  a2Dn~2  +  •  •  •  +  an)  y  =  0      (69a) 

factor  the  symbolic  operator  and  so  reduce  the  equation  to 
the  form 

(D  -  n)  (D  -  r2)  •  •  •  (D  -  rn)  y  =  0. 

The  value  y  =  Cienx  is  a  solution;  for 

(D—  ri)  cier*x  =  CineriX  —  ncie7"1*  =  0 

and  the  equation  can  be  written 

{D-ri)  •  •  •  (D-rn)  •  (D-r1)y  =  (D-r2)  •  ■  ■  (D-rn)-0  =  0. 

Similarly,  y  =  c2er2X}  y  =  c3er31,  etc.,  are  solutions.     Finally 

y  =  deriX+  c2er*x  +  •  •  •  +  cner»x  (69b) 

is  a  solution;  for  the  result  of  operating  on  y  is  the  sum  of 
the  results  of  operating  on  CieriX,  c2eT2X,  etc.,  each  of  which 
is  zero. 

If  the  roots  rh  r2,  .  .  .  ,  rn  are  all  different,  (69b)  contains 
n  constants  and  so  is  the  complete  solution  of  (69a).  If, 
however,  two  roots  r±  and  r2  are  equal 

CieriX  +  c2er*x  =  (ci  +  c2)  eTlX 


Art.  69  Equation  with  Right  Hand  Member  Zero  149 

contains  only  one  abitrary  constant  cx  +  c2  and  (69b)  con- 
tains only  n  —  1  arbitrary  constants.  In  this  case,  however, 
xeriX  is  also  a  solution;   for 

(D  —  Vi)  xenx  =  T\xeTlX  +  eTlX  —  i\xeriX  =  eriX 
and  so 

(D  -  n)2  xer*  =  (D  —  ri)  er*  =  0. 

If  then  two  roots  7*1  and  r2  are  equal,  the  part  of  the  solu- 
tion corresponding  to  these  roots  is 

(ci  +  c2x)eTlX. 

More  generally,  if  m  roots  rh  r2,  .  .  .  rm  are  equal,  the  part 
of  the  solution  corresponding  to  them  is 

(ci  +  c2x  +  czx2  +  -  •  •  +  cmxm-l)er*.  (69c) 

If  the  coefficients  ah  a2,  .  .  .  an,  are  real,  imaginary  roots 
occur  in  pairs 

ri  =  a  +  /3  V~;         r2  =  a  -  (3  V^l. 

The  terms  ci<flX,  c2er22;  are  imaginaiy  but  they  can  be  replaced 
by  two  other  terms  that  are  real.  Using  these  values  of  r\ 
and  r2)  we  have 

(D  -  n)  (D  -  r2)  =  {D-  a)2  +  /32. 

By  performing  the  differentiations  it  can  easily  be  verified 
that 

[(D  -  a)2  +  /32]  •  e<«  sin  fix  =  0, 

[(D  -  a)2  +  fi2]  •  eax  cos  #c  =  0. 

Therefore 

eax  [ci  sin  /3:r  +  c2  cos  jSx]  (69d) 

is  a  solution.     This  function,  in  which  a  and  fi  are  real,  can, 
therefore,  be  used  as  the  part  of  the  solution  corresponding 
to  two  imaginary  roots  r  =  a  db  fi  V—  1. 
To  solve  the  differential  equation 

(Z>  +  a^-1  +  a2Z>"2  +  •  •  •  +  a»)  y  -  0, 


150  Differential  Equations  Chap.  10 

let  ri,  T2,  .  .  .  ,  rn  be  the  roots  of  the  auxiliary  equation 

rn  _L.  airn-l  _|_  fl2rn-2  _|_      .    .    .     +  a„  =  0. 

//  i/iese  roois  are  aZ£  rea£  and  different,   the  solution  of  the 
equation  is 

y  =  CieriX  +  c2er*  +  •  •  •   +  cner-x. 

If  m  of  the  roots  rh  r2,  .  .  .  ,  rm  are  equal,  the  corresponding 
part  of  the  solution  is 

(ci  +  c2x  +  czx2  +  •  •  •   +  cmxm-1)  er*. 

The  part  of  the  solution  corresponding  to  two  imaginary  roots 
r  =  a  ±  (3  V  —  1  is 

eax  [ci  sin  fix  +  c2  cos  fix]. 

d2v      dii 
Example  1.     ~—-rl  —  2y  =  0. 

This  is  equivalent  to 

(D2  -  D  -  2)  y  =  0. 

The  roots  of  the  auxiliary  equation 

f-  -  r  -  2  =  0 

are  —  1  and  2.     Hence  the  solution  is 

y  =  C\C~X  +  c2c2x. 

Ex.2.    **•+**_  6* +  3y-0. 

arc3      aar         aa; 

The  roots  of  the  auxiliary  equation 

r3  +  r2  —  5r  +  3  =  0 

are  1,1,  —  3.     The  part  of  the  solution  corresponding  to  the 
two  roots  equal  to  1  is 

(ci  +  c2x)  ex. 
Hence 

y  =  (ci  +  tyx)  ex  +  c3e~3x. 


Art.  70  Right  Hand  Member  a  Function  of  x  151 

Ex.3.     (Z)2  +  2Z)  +  2)|/  =  0. 

The  roots  of  the  auxiliary  equation  are 


-  ldb  v-Hl. 

Therefore  a  =  —1,  (3  =  1  hi  (69d)  and 

y  =  e~x  [ci  sin  x  +  c2  cos  x]. 

70.  Equation  with  Right  Hand  Member  a  Function  of  x. — 
Let  y  =  u  be  the  general  solution  of  the  equation 

(Z>»  +  a.D"-1  +  a2D""2  +  •  •  •  +an)  y  =  0 

and  let  ?/  =  y  be  am/  solution  of  the  equation 

(Z>»  +  QiD"-1  +  a2Z>"2  +  •  •  •  o»)  J/  =  /  Oc).        (70) 

Then 

y  =  u  +  v 

is  a  solution  of  (70) ;  for  the  operation 

Dn  +  aj)n-1  +  a2Dn~2  +  •  •  •  +  an 

when  performed  on  u  gives  zero  and  when  performed  on  v 
gives  /  (x) .  Furthermore,  u  +  v  contains  n  arbitrary  con- 
stants.    Hence  it  is  the  general  solution  of  (70). 

The  part  u  is  called  the  complementary  function,  v  the 
'particular  integral.  To  solve  an  equation  of  the  form  (70), 
first  solve  the  equation  with  right  hand  member  zero  and 
then  add  to  the  result  any  solution  of  (70). 

A  particular  integral  can  often  be  found  by  inspection. 
If  not,  the  general  form  of  the  integral  can  usually  be  deter- 
mined by  the  following  rules: 

1.  If  /  (x)  =  axn  +  axxn~x  +  •  •  •  +  oB,  assume 

y  =  Axn  +  AiXn_1  +  •  •  •  +  An. 

But,  if  0  occurs  m  times  as  a  root  in  the  auxiliary  equation, 
assume 

y  =  xm  [Axn  +  Ai.Tn_1  +   •   •   •    +^m]. 

2.  If  /  (x)  =  ceax,  assume 

y  =  Aec 


odX 


152  Differential  Equations  Chap.  10 

But,  if  a  occurs  m  times  as  a  root  of  the  auxiliary  equation, 
assume 

y  =  Axmeax. 

3.  If  /  (#)  =  a  cos  fix  +  b  sin  fix,  assume 

y  =  A  cos  fix  +  B  sin  fix. 

But,  if  cos  fix  and  sin  fix  occur  in  the  complementary  function, 

assume 

y  =  x  [A  cos  fix  +  B  sin  fix]. 

4.  If  /  (x)  =  ae"*  cos  /to  +  beax  sin  /to,  assume 

2/  =  Aeax  cos  /to  +  fieax  sin  /to. 

But,  if  eax  cos  /to  and  eaz  sin  fix  occur  in  the  complementary 
function,  assume 

y  =  xe™  [A  cos  fix  +  B  sin  /to]. 

If  /  (x)  contains  terms  of  different  types,  take  for  y  the 
sum  of  the  corresponding  expressions.  Substitute  the  value 
of  y  in  the  differential  equation  and  determine  the  constants 
so  that  the  equation  is  satisfied. 

Example  1.     -y{  +  ±y  =  2x +  Z. 


dx2 
A  particular  solution  is  evidently 

y  =  i  (2x  +  3). 

Hence  the  complete  solution  is 

y  =  Ci  cos  2  x  +  c2  sin  2  a;  +  J  (2  £  +  3). 

Ex.  2.     (D2  +  3  D  +  2)  2/  =  2  +  e*. 
Substituting  ?/  =  A  +  ite*,  we  get 

2  A  +  6Bex  =  2  +  ex. 
Hence 

2A  =  2,        65=1 
and 

2/  =  1  +  J  ez  +  ci<rx  -f  c2e-2r. 

£x.3.     ?3  +  f{  =  *. 

ax*      ax2 


Art.  71  Simultaneous  Equations  153 

The  roots  of  the  auxiliary  equation  are  0,  0,  —  1.  Since 
0  is  twice  a  root,  we  assume 

y  =  x2  (Ax2  +  Bx  +  C)  =  Ax"  +  Bxz  +  Cx2. 

Substituting  this  value, 

12  Ax2  +  (24  A  +  Q>B)x  +  6B  +  2C  =  x2. 

Consequently, 

12A  =  1,        24A+6£  =  0,        6J3  +  2C  =  0, 
whence 

A=TV,  B=-l        C--1. 

The  solution  is 

V  =  tV  z4  —  -|  x3  +  x2  +  ci  +  c2a;  +  c3e~x. 

71.  Simultaneous  Equations.  —  We  consider  only  linear 
equations  with  constant  coefficients  containing  one  independ- 
ent variable  and  as  many  dependent  variables  as  equations. 
All  but  one  of  the  dependent  variables  can  be  eliminated  by 
a  process  analogous  to  that  used  in  solving  linear  algebraic 
equations.  The  one  remaining  dependent  variable  is  the 
solution  of  a  linear  equation.  Its  value  can  be  found  and 
the  other  functions  can  then  be  determined  by  substituting 
this  value  in  the  previous  equations. 

Example.  -r-  -\-2x  —  3y  =  t, 

Using  D  for  -r ,  these  equations  can  be  written 

(D  +  2)x-3y  =  t, 
(D  +  2)y-  Sx  =  e2t. 

To  eliminate  y,  multiply  the  first  equation  by  D  +  2  and  the 
second  by  3.     The  result  is 

(D  +  2)2x-S(D+2)y=  1+2*, 
3(D+2)i/-9x  =  de2t. 


154  Differential  Equations  Chap.  10 

Adding,  we  get 

[(D  +  2Y-9]x=  l+2t+3e2t. 
The  solution  of  this  equation  is 

x  =  -  1 1  -  U  +  f  e««+  Cie«  +  c2e-5'. 
Substituting  this  value  in  the  first  equation,  we  find 
y  =  i(D  +  2)  s  -  i  t  =  -  §  t  -  U  +  *  e2*  +  c^  -  c2e"5t. 

EXERCISES 

Solve  the  following  equations: 
.,        d%  ,  dy   ,  „ 

2.    (x  +  l)g-(x  +  2)|  +  x  +  2  =  0. 

o    d2y        9  i»    d2y  ,  dy   ,  rt 

4    &y=-a2v  17    <Vy_od*y 

8- *  ^  - *  +  v^  j  • 

9.   ^_4^  =  o 
drc2         dx 

»S-»g-«.-a  23.g-2|  =  2*-3. 

12.   g  +  2/  =  0.  *g'-*-*". 

13- a5-2d5-3d5  =  0-        26- ^-s+y  =  cos2x- 

»3-4+«»-'*        •3-431+1,-*.*.* 


20. 

d2?/ 

£-*»-'• 

21. 

S  +  E-"* 

22. 

dw 

Art.  71                          Simultaneous  Equations 

d2v 
29.   -A  —  9y  =  e3x  cos x. 

dx2 

31.   ^  +  4j/  =  cos2x. 

nn    dfy  ,  d3y             . 

30-  di+di°  =  cosix- 

32.   g  +  a4  +  f_, 

dx2         dx       u 

33.   %  +  ,  =  e<, 

dx                  * 

34.    ft=x-2y  +  l, 

3 —.*  +  »■ 

oc      .  cte      dy   ,   r, 
35.    4  —  -  -57  +  3  x  =  sin  *, 
dt       dt 

36     ^-z 
6b.    dt2  -  x, 

37.    Solve  the  equation 

djy 

g)2  = J 

155 


dv 
and  determine  the  constants  so  that  y  =  0  and  ■—  =  1  when  x  =  0. 

70  7 

38.  Solve  ~  =  3  Vy  under  the  hypothesis  that  ?/  =  1  and  -~  =  2 

when  z  =  0. 

39.  When  a  body  sinks  slowly  in  a  liquid,  its  acceleration  and 
velocity  approximately  satisfy  the  equation 

a  =  g  —  kv, 

g  and  k  being  constants.     Find  the  distance  passed  over  as  a  function 
of  the  time  if  the  body  starts  from  rest. 

40.  The  acceleration  and  velocity  of  a  body  falling  in  the  air  approxi- 
mately satisfy  the  equation  a  =  g  —  kv2,  g  and  k  being  constants. 
Find  the  distance  traversed  as  a  function  of  the  time  if  the  body  falls 
from  rest. 

41.  A  weight  supported  by  a  spiral  spring  is  lifted  a  distance  b  and 
let  fall.  Its  acceleration  is  given  by  the  equation  a  =  —  k2s,  k  being 
constant  and  s  the  displacement  from  the  position  of  equilibrium. 
Find  s  in  terms  of  the  time  t. 

42.  Find  the  velocity  with  which  a  meteor  strikes  the  earth,  assum- 
ing that  it  starts  from  rest  at  an  indefinitely  great  distance  and  moves 
toward  the  earth  with  an  acceleration  inversely  proportional  to  the 
square  of  its  distance  from  the  center. 

43.  A  body  falling  in  a  hole  through  the  center  of  the  earth  would 
have  an  acceleration  toward  the  center  proportional  to  its  distance  from 
the  center.  If  the  body  starts  from  rest  at  the  surface,  find  the  time 
required  to  fall  through. 


156  Differential  Equations  Chap.  10 

44.  A  chain  5  feet  long  starts  with  one  foot  of  its  length  hanging 
over  the  edge  of  a  smooth  table.  The  acceleration  of  the  chain  will  be 
proportional  to  the  amount  over  the  edge.  Find  the  time  required  to 
slide  off. 

45.  A  chain  hangs  over  a  smooth  peg,  8  feet  of  its  length  being  on 
one  side  and  10  on  the  other.  Its  acceleration  will  be  proportional  to 
the  difference  in  length  of  the  two  sides.  Find  the  time  required  to 
slide  off. 


SUPPLEMENTARY  EXERCISES 


/ 


.r  dx 


a  +  bx2 
[  (a  +  bx)2  dx. 


1. 

2. 

3. 

p  +  gx 

4.    fxV2  -  3x2dx. 
(x  +  a)  dx 


s 


a  +  bx 


dx. 


5. 

6. 

7. 

8. 

9. 

.0. 

11. 

12. 

[3. 

14. 


J  Vz2  +  2  ax  +  6 

J     J         X  X2 

C(x  -  1)  (x2  -2x)*  dx. 

/dx 
i 


sin  ax 


f  sin  ax    , 
I  — ; —  dx. 
J  cos2  ax 


/cc 


cos  2x 


sin  2x 

/sec2  x  tan  x  dx 
a  -{-  b  sec2  a; 
dx 


dx. 


/ 

/cot 


sec  a: 
cot  a:  dx 
sin  x 
i  ax 


cos  ox 
dx 


dx. 


6.     f- 

J    I 

17-  /__, 


cos  x  —  sin  x 
dx 


sec  x  —  tan  x 

dx 
sin2  ax  cos2  ax 


CHAPTER   II 

18. 
19. 

20. 

21. 
22. 

23. 
24. 
25. 
26. 
27. 
28. 
29. 
30. 
31. 
32. 
33. 

34. 
157 


cos  x  dx 

-  I,- 1 

cos  x  +  sin  x 
dx 

— --■  ii  • 

Vtan2  x  +  2 

dx 
X  Vxn  —  1 
vV-  -  1 


dx. 


dx 


sin2  x  —  cos2  x 
cot  x  dx 

— —  • 

Vl  +  sin2  x 
dx 


/ 
/ 

/ 

J  (2x--  1)  V4X2" 
J  xeax*  dx. 

b  +  ce^' 
j  sec2  xe**11  x  dx. 

Cah+ex  dx. 

/dx 

C    dx    . 

Je1-  e~x 
I  tan  x  In  cos  x  a'x. 

/dx 
a! 

/ 


4x 


2x2  +  2  abx  +  b2 
x  dx 

Vx2  -  2  x  +  3 
(2  x  +  3)  dx 


(x  -  1)  Vx2  -  2 


X 


158  Supplementary  Exercises 

35-  fTvim'  56-  f*w^^2d*- 

36.  f    j; .  57.  <•-**=. 

37.  f(a*-a?)*dx.  58.   /^jji" 

38.  f      /*         .  59.     f— ** .. 

39.  fV3  -2x-x*dx.  60-     f  /,   ^     • 
J  J  v  a3  —  x3 

40.  f  (a? -a^dz.  61.     {(?  +  ***  dx. 

41.  j  cos6 re  sin2  x  dx.  62.     J  e0*  sin  6x  d#. 

42.  f(l+cosx)*dx.  63.     f ^  dr. 

43.  J  tan  2  z  sec5  2  x  dx.  64.     J  a;  In  (ex  +  &)  cZaj. 

44.  fcoVxdx.  65.   Jln  (a*  +  b)  dx 

45-    /tan x  +  cot/  '  66-   /*<**-**&>. 

46.  f  (seca:  +  tana;)2da\  67      f xdx 

J  J  (x2  -  l)2  (a;2  +  1) 

47.  f tan  *  ~  j  da;.  68      f x2dY 

J  tana;  +  1  uo-    J  (;C3  +  !)  (Xs  _  2)  * 

.„       r cos3 xdx  r   t rl-r 

48-   J  "S?i-  69.   /^fj. 

49.     I  sin  2  x  cos2  a:  dx.  70      f      ^         ^ 
J  J  (x  —  l)5 


50. 


JVl+cos2a;sin2a;da;.  71     JV  cos  1  ^a;. 


55. 


51.     f-    /dX  72      f  2a;2  +  3a; 

J  x  Va2x  +  o2  •    J  (x-i)  (x-2)  (X+S) 

52      f ^ 70      f(3s-5)<fo 


<ia;. 


x2  Vx  -  2  73"   J    a;  (a;  +  3)2 

53.  f £==-  74.     f-^-. 

J  (x  -  1)  Vx  +  2  J  a;3  -  8 

54.  fx  (ax  +  6)*  da;.  75.     f        x       ■ 

»/  -/  (1  —  a:3)3 


(1  -  re3)' 
a:3  da; 


r^x*,.  76.  r_^! 


Voa;  +  6  «/  (1  -  2a;2)f 


Supplementary  Exercises  159 

x2  -  3  x  +  2 


77.  f sec4  x  tan*  x  dx.  84.     f 
^  »/  Vx-  -    4  x  ■  -  3 

78.  [  sin  3  x cos  4 x dx.  g_      fx/^ 


dx. 


—  x 


+  x 


79 


•   /■* *<*>b2 zdx.  86>  J(sinx_COsx)3^. 

80.  |  sin  x  sin  5  x  dx.  »7  f ^^     . 

J  J  (x2  -a2)2 

81.  fcos  2  x  cos  3  x  dx.  gg  flog  (x  +  Vx2  -  l) 

J  Vx2  -  1 

82.  (   (cot  X  +  CSC  x)2  C?X.  on  f        7      j 

J  89.  I  sec7  x  ax. 

Q        f    (3x  -  l)dx  r 

83-    JV2  +  3X-/  90-  J>  +  «2)*^ 


CHAPTER  IV 

91.  Find  the  area  bounded  by  the  x-axis  and  the  parabola  y  =  x2 
-  4x  +  5. 

92.  Find  the  area  bounded  by  the  curves  y  =  x3,  y2  =  x. 

93.  Find  the  area  bounded  by  the  parabola  y2  =  2  x  and  the  witch 
1_ 

X~  y2  +  l 

94.  Find  the  area  within  a  loop  of  the  curve  y2  =  x2  —  x4. 

95.  Find  the  area  of  one  of  the  sectors  bounded  by  the  hyperbola 
x2  —  y2  =  3  and  the  lines  x  =  ±  2y. 

96.  Find  the  area  bounded  by  the  parabolas  y2  =  2  ax  +  a2,  y2  + 
2  ax  =  0. 

97.  Find  the  area  within  the  loop  of  the  curve  x  =  :; — ; -,  y  = 

1  +  mz 

3  am2 


1  +m3 

98.  Find  the   area  bounded  by  the  parabola  x  =  a  cos  2  <f>,  y  = 
a  sin  0  and  the  line  x  =  —  a. 

99.  Find  the  area  inclosed  by  the  curve  x  =  a  cos3  <f>,  y  =  b  sin3  </>. 

100.  Find  the  area  bounded  by  the  curve  x  =  asind,  y  =  a  cos8  d. 

101.  Find  the  area  of  one  loop  of  the  curve  r  =  a  cos  nd. 

102.  Find  the  area  of  a  loop  of  the  curve  r  =  a  (1  —  2  cos  0). 

103.  Find  the  area  between  the  curves  r  =  a  (cos  0  +  2),  r  =  a. 

104.  Find  the  total  area  inclosed  by  the  curve  r  =  a  sin  £  0. 

105.  Find  the  area  of  the  part  of  one  loop  of  the  curve  r2  =  a2  sin  3  0 
outside  the  curve  r2  =  a  sin  0. 


160  Supplementary  Exercises 

106.  By  changing  to  polar  coordinates  find  the  area  within  one  loop 
of  the  curve  (x2  +  y2)2  =  a?xy. 

107.  By  changing  to  polar  coordinates  find  the  area  of  one  of  the 
regions  between  the  circle  x2  +  y2  =  2  a2  and  hyperbola  x2  —  y2  =  a2. 

108.  Find  the  area  of  one  of  the  regions  bounded  by  d  =  sin  r  and  the 
line  0  =  1. 

109.  Find  the  volume  generated  by  revolving  an  ellipse  about  the 
tangent  at  one  of  its  vertices. 

110.  Find  the  volume  generated  by  revolving  about  the  ?/-axis  the 
area  bounded  by  the  curve  y2  =  xz  and  the  line  x  =  4. 

111.  Find  the  volume  generated  by  rotating  about  the  y-axis  the  area 
between  the  z-axis  and  one  arch  of  the  cycloid  x  =  a  (<£  —  sin  <£), 
y  =  a  (1  —  cos^). 

112.  Find  the  volume  generated  by  rotating  the  area  of  the  preced- 
ing problem  about  the  tangent  at  the  highest  point  of  the  cycloid. 

113.  Find  the  volume  generated  by  revolving  about  the  x-axis  the 
part  of  the  ellipse  x2  —  xy  +  y2  =  1  in  the  first  quadrant. 

114.  Find  the  volume  generated  by  revolving  about  0  =  ^  the  area 

enclosed  by  the  curve  r2  =  a2  sin  0. 

115.  The  ends  of  an  ellipse  move  along  the  parabolas  z2  =  ax, 
y2  =  ax  and  its  plane  is  perpendicular  to  the  x-axis.  Find  the  volume 
swept  out  between  x  =  0  and  x  =  c. 

116.  The  ends  of  a  helical  spring  lie  in  parallel  planes  at  distance  h 
apart  and  the  area  of  a  cross  section  of  the  spring  perpendicular  to  its 
axis  is  A.     Find  the  volume  of  the  spring. 

117.  The  axes  of  two  right  circular  cylinders  of  equal  radius  intersect 
at  an  angle  a.     Find  the  common  volume. 

118.  A  rectangle  moves  from  a  fixed  point,  one  side  varying  as  the 
distance  from  the  point,  and  the  other  as  the  square  of  this  distance. 
At  the  distance  of  10  feet  the  rectangle  becomes  a  square  of  side  4  ft. 
What  is  the  volume  then  generated? 

119.  A  cylindrical  bucket  filled  with  oil  is  tipped  until  half  the  bot- 
tom is  exposed;  if  the  radius  is  4  inches  and  the  altitude  12  inches  find 
the  amount  of  oil  poured  out. 

120.  Two  equal  ellipses  with  semi-axes  5  and  6  inches  have  the  same 
major  axis  and  lie  in  perpendicular  planes.  A  square  moves  with  its 
center  in  the  common  axis  and  its  diagonals  chords  of  the  ellipses. 
Find  the  volume  generated. 

121.  Find  the  volume  bounded  by  the  paraboloid  12  z  =  3  x2  +  y2 
and  the  plane  z  =  4. 


Supplementary  Exercises  161 

CHAPTER  V 

122.  Find  the  length  of  the  arc  of  the  curve 

y  =  i  x  V.r-  —  1  —  i  In  (x  +  Vx*  —  l)  between  x  =  1  and  x  =  3. 

123.  Find  the  arc  of  the  curve  9  if  =  (2  x  —  l)3  cut  off  by  the  line 
x  =  5. 

124.  Find  the  perimeter  of  the  loop  of  the  curve 

9x2  =  {2y-l)(y-2)2. 

125.  Find  the  length  of  the  curve  x  =  t2  +  t,  y  =  I2  —  t  below  the 
x-axis. 

126.  Find  the  length  of  an  arch  of  the  curve 

x  =  aV3(2</>  —  sin  2  0),     y  =  ^  (1  -  cos  3  0). 

127.  Find  the  length  of  one  quadrant  of  the  curve 

x  =  a  cos3  </>,         y  =  b  sin3  <f>. 

128.  Find  the  circumference  of  the  circle 

r  =  2  sin  6  +  3  cos  6. 

129.  Find  the  perimeter  of  one  loop  of  the  curve 


r  =  a  sin5 


(?) 


130.  Find  the  area  of  the  surface  generated  by  revolving  the  arc  of 
the  curve  9  y2  =  (2  x  —  l)3  between  x  =  0  and  x  =  2  about  the  y-axis. 

131.  Find  the  area  of  the  surface  generated  by  revolving  one  arch  of 
the  cycloid  x  =  a  (<f>  —  sin  0),  y  =  a(l  —  cos  0)  about  the  tangent  at 
its  highest  point. 

132.  Find  the  area  of  the  surface  generated  by  rotating  the  curve 
r2  =  a2  sin  2  6  about  the  x-axis. 

133.  Find  the  area  generated  by  revolving  the  loop  of  the  curve 
9  x2  =  (2  y  —  1)  (y  —  2)a  about  the  x-axis. 

134.  Find  the  volume  generated  by  revolving  the  area  within  the 
curve  y2  =  x2  (1  —  x2)  about  the  y-axis. 

135.  The  vertical  angle  of  a  cone  is  90°,  its  vertex  is  on  a  sphere  of 
radius  a,  and  its  axis  is  tangent  to  the  sphere.  Find  the  area  of  the  cone 
within  the  sphere. 

136.  A  cylinder  with  radius  b  intersects  and  is  tangent  to  a  sphere  of 
radius  a,  greater  than  b.  Find  the  area  of  the  surface  of  the  cylinder 
within  the  sphere. 

137.  A  plane  passes  through  the  center  of  the  base  of  a  right  circular 
cone  and  is  parallel  to  an  element  of  the  cone.  Find  the  areas  of  the 
two  parts  into  which  it  cuts  the  lateral  surface. 


162  Supplementary  Exercises 

CHAPTER   VI 

138.  Find  the  pressure  on  a  square  of  side  4  feet  if  one  diagonal  is 
vertical  and  has  its  upper  end  in  the  surface. 

139.  Find  the  pressure  on  a  segment  of  a  parabola  of  base  2  b  and 
altitude  h,  if  the  vertex  is  at  the  surface  and  the  axis  of  the  parabola  is 
vertical. 

140.  Find  the  pressure  on  the  parabolic  segment  of  the  preceding 
problem  if  the  vertex  is  submerged  and  the  base  of  the  segment  is  in  the 
surface. 

141.  Find  the  pressure  on  the  ends  of  a  cylindrical  tank  4  feet  in 
diameter,  if  the  axis  is  horizontal  and  the  tank  is  filled  with  water  under 
a  pressure  of  10  lbs.  per  square  inch  at  the  top  of  the  tank. 

142.  A  barrel  3  ft.  in  diameter  is  filled  with  equal  parts  of  water  and 
oil.  If  the  axis  is  horizontal  and  the  weight  of  oil  half  that  of  water, 
find  the  pressure  on  one  end. 

143.  Find  the  moment  of  the  pressure  in  Ex.  138  about  the  other 
diagonal  of  the  square. 

144.  Weights  of  1,  2,  and  3  pounds  are  placed  at  the  points  (0,  0), 
(2,  1),  (4,  —  3).     Find  their  center  of  gravity. 

145.  A  trapezoid  is  formed  by  connecting  one  vertex  of  a  rectangle 
to  the  middle  point  of  the  opposite  side.     Find  its  center  of  gravity. 

146.  Find  the  center  of  gravity  of  a  sector  of  a  circle  with  radius  a 
and  central  angle  2  a. 

147.  Find  the  center  of  gravity  of  the  area  within  a  loop  of  the  curve 
y2  =  x2  —  xi. 

148.  Find  the  center  of  gravity  of  the  area  bounded  by  the  curve 

•v»3 

y2  =  — and  its  asymptote  x  =  2  a. 

149.  Find  the  center  of  gravity  of  the  area  within  one  loop  of  the 
curve  r2  =  a2  sin  0. 

150.  Find  the  center  of  gravity  of  the  area  of  the  curve  x  =  a  sin3  <j>, 
y  =  b  sin3  4>  above  the  x-axis. 

151.  Find  the  center  of  gravity  of  the  arc  of  the  curve  9  y2  = 
(2  x  —  l)3  cut  off  by  the  line  x  =  5. 

152.  Find  the  center  of  gravity  of  the  arc  that  forms  the  loop  of  the 

curve 

9  y2  =  (2  x  -  1)  (x  -  2)2. 

153.  Find  the  center  of  gravity  of  the  arc  of  the  curve  x  =  t2  +  t, 
y  =  t2  —  t  below  the  x-axis. 

154.  Show  that  the  center  of  gravity  of  a  pyramid  of  constant 
density  is  on  the  line  joining  the  vertex  to  the  center  of  gravity  of  the 
base,  f  of  the  way  from  the  vertex  to  the  base. 


Supplementary  Exercises  1G3 

155.    Find  the  center  of  gravity  of  the  surface  of  a  right  circular  cone. 

15G.  Show  that  the  distance  from  the  base  to  the  center  of  gravity 
of  the  surface  of  an  oblique  cone  is  3  of  the  altitude.  Is  it  on  the  line 
joining  the  vertex  to  the  center  of  the  base? 

157.  Find  the  center  of  gravity  of  the  solid  generated  by  rotating 
about  the  line  x  =  4,  the  area  above  the  rc-axis  bounded  by  the  parab- 
ola if-  =  4  x  and  the  line  x  =  4. 

158.  The  arc  of  the  curve  x*  -f-  y '  =  a*  above  the  z-axis  is  rotated 
about  the  ?/-axis.  Find  the  center  of  gravity  of  the  volume  and  that 
of  the  area  generated. 

159.  Assuming  that  the  specific  gravity  of  sea  water  at  depth  h  in 

miles  is 

P  =  e-007'0  h, 

find  the  center  of  gravity  of  a  section  of  the  water  with  vertical  sides 
five  miles  deep. 

160.  By  using  Pappus's  theorems,  find  the  center  of  gravity  of  the 
arc  of  a  semicircle. 

161.  The  ellipse 

a2      b2 

is  rotated  about  a  tangent  inclined  45°  to  its  axis.     Find  the  volume 
generated. 

162.  The  volume  of  the  ellipsoid 

t.  _|_  t.   _L  £    =    1 

is  f  irabc.     Use  this  to  find  the  center  of  gravity  of  a  quadrant  of  the 

x2      y2 
ellipse  ^-1-^=  1- 

163.  Find  the  volume  generated  by  revolving  one  loop  of  the  curve 
r  =  a  sin  0  about  the  initial  line. 

164.  A  semicircle  of  radius  a  rotates  about  its  bounding  diameter 
while  the  diameter  slides  along  the  line  in  which  it  lies.  Find  the 
volume  generated  in  one  revolution. 

165.  The  plane  of  a  moving  square  is  perpendicular  to  that  of  a  fixed 
circle.  One  corner  of  the  square  is  kept  fixed  at  a  point  of  the  circle 
while  the  opposite  corner  moves  around  the  circle.  Find  the  volume 
generated. 

166.  Find  the  moment  of  inertia  about  the  z-axis  of  the  area  bounded 
by  the  .r-axis  and  the  curve  y  =  4  —  x2. 

167.  Show  that  the  moment  of  inertia  of  a  plane  area  about  an  axis' 
perpendicular  to  its  plain'  .it  the  origin  is  equal  to  the  sum  of  its  moments 
of  inertia  about  the  coordinate  axes.     Use  this  to  find  the  moment  of 


164  Supplementary  Exercises 

o*2  o  /2 

inertia  of  the  ellipse  —  +  rz  —  1  about  the  axis  perpendicular  to  its 

plane  at  its  center. 

168.  Find  the  moment  of  inertia  of  the  surface  of  a  right  circular  cone 
about  its  axis. 

169.  The  area  bounded  by  the  x-axis  and  the  parabola  y2  =  4  ax  —  x2 
is  revolved  about  the  z-axis.  Find  the  moment  of  inertia  about  the 
rc-axis  of  the  volume  thus  generated. 

170.  From  a  right  circular  cylinder  a  right  cone  with  the  same  base 
and  altitude  is  cut.  Find  the  moment  of  inertia  of  the  remaining 
volume  about  the  axis  of  the  cylinder. 

171.  A  torus  is  generated  by  rotating  a  circle  of  radius  a  about  an 
axis  in  its  plane  at  distance  b,  greater  than  a,  from  the  center.  Find 
the  momeat  of  inertia  of  the  volume  of  the  torus  about  its  axis. 

172.  Find  the  moment  of  inertia  of  the  area  of  the  torus  about  its  axis. 

173.  The  kinetic  energy  of  a  moving  mass  is 


J  \  v2  dm, 


where  v  is  the  velocity  of  the  element  of  mass  dm.  Show  that  the 
kinetic  energy  of  a  homogeneous  cylinder  of  mass  M  and  radius  a 
rotating  with  angular  velocity  w  about  its  axis  is  §  Mccra1. 

174.  Show  that  the  kinetic  energy  of  a  uniform  sphere  of  mass  M  and 
radius  a  rotating  with  angular  velocity  00  about  a  diameter  is  \  Moo2a2. 

175.  When  a  gas  expands  without  receiving  or  giving  out  heat,  its 
pressure  and  volume  are  connected  by  the  equation 

pvv  =  k 

where  7  and  k  are  constant.  Find  the  work  done  in  expanding  from 
the  volume  V\  to  the  volume  v2  . 

176.  The  work  done  by  an  electric  current  of  i  amperes  and  E  volts 
is  iE  joules  per  second.     If 

E  =  E0  cos  cot,        i  —  Io  cos  (cot  +  «), 

where  Eo,  I0,  w  are  constants,  find  the  work  done  in  one  cycle. 

177.  When  water  is  pumped  from  one  vessel  into  another  at  a  higher 
level,  show  that  the  work  in  foot  pounds  required  is  equal  to  the  product 
of  the  total  weight  of  water  in  pounds  and  the  distance  in  feet  its  center 
of  gravity  is  raised. 

CHAPTER  VII 

178.  Find  the  volume  of  an  ellipsoid  by  using  the  prismoidal  formula. 

179.  A  wedge  is  cut  from  a  right  circular  cylinder  by  a  plane  which 
passes  through  the  center  of  the  base  and  makes  with  the  base  an  angle  a. 
Find  the  volume  of  the  wedge  by  the  prismoidal  formula. 


Supplementary  Exercises  165 

180.  Find  approximately  the  volume  of  a  barrel  30  inches  long  if  its 
diameter  at  the  ends  is  20  inches  and  at  the  middle  24  inches. 

181.  The  width  of  an  irregular  piece  of  land  was  measured  at  inter- 
vals of  10  yards,  the  measurements  being  52,  5G,  67,  49,  45,  53,  and  62 
yards.     Find  its  area  approximately  by  using  Simpson's  rule. 

Find  the  values  of  the  following  integrals  approximately  by  Simpson's 
rule: 

f  Vx3  +  1  dx.  184.     f   * Vsin  x  dx. 

Jo  Jo 

Cl°-Jnxdx.  185.     f\-^-v 

Jl      X2  J_4  1   +  X4 


1S2. 
183. 


186.  Find  approximately  the  length  of  an  arch  of  the  curve  y  =  sin  x. 

187.  Find  approximately  the  area  bounded  by  the  x-axis,  the  curve 


y  =  ,  and  the  ordinates  x  =  0,  x  =  ir. 

x 


CHAPTER  VIII 

Express  the  following  quantities  as  double  integrals  and  determine 
the  limits: 

188.  Area  bounded  by  the  parabola  y  =  x2  —  2rc  +  3  and  the  line 
y  =  2x. 

189.  Area  bounded  by  the  circle  x2  +  y2  =  2  a2  and  the  curve 

y2  = 


2a  —  x 

190.  Moment  of  inertia  about  the  a>axis  of  the  area  within  the  circles 

x2  +y2  =b,         x2  +  if  -  2  x  -  4  y  =  0. 

191.  Moment  of  inertia  of  the  area  within  the  loop  of  the  curve 
y2  =  x2  —  x*  about  the  axis  perpendicular  to  its  plane  at  the  origin. 

192.  Volume  bounded  by  the  xy-p\ane  the  paraboloid  z  =  x2  +  y2 
and  the  cylinder  x2  +  y2  =  4. 

193.  Volume  bounded  by  the  zy-plane  the  paraboloid  z  =  x2  +  y2 
and  the  plane  z  =  2  x  +  2  y. 

194.  Center  of  gravity  of  the  solid  bounded  by  the  zz-plane,  the 
cylinder  x2  -+-  z2  =  a2,  and  the  plane  x  +  y  +  z  =  4  a. 

195.  Volume  generated  by  rotating  about  the  z-axis  one  of  the  areas 
bounded  by  the  circle  x1  -\-  y2  =  5  a2  and  the  parabola  y2  =  4  ax. 

In  each  of  the  following  cases  determine  the  region  over  which  the 
integral  is  taken,  interchange  dx  and  dy,  determine  the  new  limits,  and 
so  find  the  value  of  the  integral: 


166  Supplementary  Exercises 

•<™       Cl    Cx  xdxdy  /»i  pi     1    —  - 

196.     J      I  JL-  198.     J     f  _  -c    «dj/dB. 

I       . (x+y)dydx.         199.     f     |     V  z2  +  x?/ di/ dz. 

0   ./a—  v a2— !/2  ^ o    •'0 

Express   the   following   quantities   as   double   integrals   using   polar 
coordinates: 

200.  Area  within  the   cardioid  r  =  a  (1  +  cos  d)   and    outside    the 
circle  r  =  f  a. 

201.  Center  of    gravity   of   the   area  within  the   circle  r  =  a  and 
outside  the  circle  r  =2a  sin  9. 

202.  Moment  of  inertia  of  the  area  cut  from  the  parabola 

2a 
r  = 

1  —  COS0 

by  the  line  y  =  x,  about  the  z-axis. 

203.  Volume  within  the  cylinder  r  =  2  a  sin  6  and  the  sphere 

x2  +  y2  +  z1  =  4  a2. 

204.  Moment  of  inertia  of  a  sphere  about  a  tangent  line. 

205.  Volume  bounded  by  the  paraboloid  z  =  x2  +  y2  and  the  plane 
z  =  2x  +  2y. 

206.  Find  the  area  cut  from  the  cone  x2  +  y2  =  z2  by  the  plane 
x  =  2z  -  3. 

207.  Find  the  area  cut  from  the  plane  by  the  cone  in  Ex.  206. 

208.  Find  the  area  of  the  surface  z2  +  (x  +  y)2  =  a2  in  the  first 
octant. 

209.  Determine  the  area  of  the  surface  z2  =  2  x  cut  out  by  the  planes 
y  =  0,  y  =  x,  x  =  1. 

CHAPTER   IX 

Express  the  following  quantities  as  triple  integrals: 

210.  Volume  of  an  octant  of  a  sphere  of  radius  a. 

211.  Moment  of  inertia  of  the  volume  in  the  first  octant  bounded  by 

x       II       z 
the  plane  -  +  r  +  -  =  1  about  the  rc-axis. 
a      b      c 

212.  Center  of  gravity  of  the  region  in  the  first  octant  bounded  by 
the  paraboloid  z  =  xy  and  the  cylinder  x2  +  y2  =  a2. 

213.  Moment  of  inertia  about  the  2-axis  of  the  volume  bounded  by 
the  paraboloid  z  =  x2  +  y2  and  the  plane  z  =  2  z  +  3. 

214.  Volume  bounded  by  the  cone  x2  =  y2  +  2  z2  and  the    plane 
3  x  +  y  =  6. 

Express  the  following  quantities  as  triple  integrals  in  rectangular, 
cylindrical,  and  spherical  coordinates,  and  evaluate  one  of  the  integrals'. 


Supplementary  Exercises  167 

215.  Moment  of  inertia  of  a  right  circular  cylinder  about  a  line 
tangent  to  its  base. 

216.  Moment  of  inertia  of  a  segment  cut  from  a  sphere  by  a  plane, 
about  a  diameter  parallel  to  that  plane. 

217.  Center  of  gravity  of  a  right  circular  cone  whose  density  varies 
as  the  distance  from  the  center  of  the  base. 

218.  Volume  bounded  by  the  xy-plane,  the  cylinder  x-  +  V2  =  2  ax 
and  the  cone  z-  =  x2  +  y1- 

219.  Find  the  attraction  of  a  unifprm  wire  of  length  I  and  mass  M 
on  a  particle  of  unit  mass  at  distance  c  from  the  wire  in  the  perpen- 
dicular at  one  end. 

220.  Find  the  attraction  of  a  right  circular  cylinder  on  a  particle  at 
the  middle  of  its  base. 

221.  Show  that  the  attraction  of  a  homogeneous  shell  bounded  by 
two  concentric  spherical  surfaces  on  a  particle  in  the  enclosed  space 
is  zero. 

CHAPTER  X 

Solve  the  following  differential  equations: 

222.  ydx  +  (x  -  xy)dy  =  0. 

223.  sin  x  sin  y  dx  +  cos  x  cos  ydy  =  0. 

224.  (2  xy  -  y2  +  6  x2)  dx  +  (3  y2  +  x2  -  2  xy)  dy  =  0. 

225.  x^+y  =  x3y. 

ax 

226.  x  -^  +  y  =  cot  x. 
dx 


227.   xdy  -  \y  +  ex )  dx  = 


0. 


228.  (1  +  x2)  dy  +  (xy  +  x)  dx  =  0. 

229.  xdx  +  y dy  =  xdy  —  y dx. 

230.  (sin  x  -\-  y)  dy  -\-  (y  cos  x  —  x2)  dx  =  0. 

231.  y  (ex  +  2)  dx  +  (ex  +  2  x)  dy  =  0. 

232.  (xy2  -  x)  dx  +  (y  +  xy)  dy  =  0. 

233.  (l+x2)^c+xy  =  2y. 


234.  x  dy  —  y  dx  =  Vx2  -\-  y2  dx. 

235.  (x  -  y)  dx  +  xdy  =0. 

236.  xdy  -  ydx  =  x  ^x2  -f  y2 dx. 

237.  e***  tfy  +  (1  +  ev)  dx  =  0. 

238.  (2  x  +  3  ?/  -  1)  dx  +  (4  x  +  6  ?/  -  5)  dy  =  0. 

239.  (3  ?/2  +  3  xy  +  x2)  dx  =  (x2  +  2  xy)  dy. 

240.  (1  +  x2)  dy  +  (xy  -  x2)  dx  =  0. 


168  Supplementary  Exercises 

241.  (x2y  +  yi)dx-  (x3  +  2  xy3)  dy  =  0. 

242.  (y  +  l)(fj  =x*-x>. 

243.  2^  +  y  +  xy*  =  0. 

244.  ydx  =  (y3  —  x)  dy. 

245.  y  -~  +  y2  cot  x  =  cos  x. 

246.  (x2  -  y2)  (dx  +  dy)  =  (xs  +  2/2)  (dy  -  dx). 

2*8-  I  =  y 

a*- cl+*)S(=l• 
25l•i=w^W• 
262■^+2g-3*=*■ 

253.  ft  -  *  -  e*». 

dx3      dx 

255-2>-id£+8y  =  3x-i- 

256.   ft  +  a^-.  +  l. 

ax3  dx 

957    ^4-2^-^-277-^4-3 
^57-   dx3  +  ^dx2      dx       22/_e    +d* 

258.  3-|  +  a2w  =  sin  ax. 
ax4 

259.  ^-^-27/  =  e*sin2x. 
dx2      dx    . 

260.  ^-^-22/  =  e-^sin2x. 

261.  t^  +  9  ?/  =  2  cos  3  x  -  3  cos  2  s. 


Supplementary  Exercises  1G9 

263.   g  -  „  -  *»- 

«•«•  £  +  »£  +  «» -.■«■* 

265.   |j[  +  2*  =  sin  (,  ^  -2y  =  cos*. 

267.  According  to  Newton's  Law,  the  rate  at  which  a  substance  cools 
in  air  is  proportional  to  the  difference  of  the  temperature  of  the  sub- 
stance and  the  temperature  of  air.  If  the  temperature  of  air  is  20°  C. 
and  the  substance  cools  from  100°  to  60°  in  20  minutes,  when  will  its 
temperature  become  30°? 

268.  A  particle  moves  in  a  straight  line  from  a  distance  a  towards  a 
point  with  an  acceleration  which  at  distance  r  from  the  point  is  k  r~ §. 
If  the  particle  starts  from  rest,  how  long  will  be  the  time  before  it 
reaches  the  point? 

269.  A  substance  is  undergoing  transformation  into  another  at  a 
rate  proportional  to  the  amount  of  the  substance  remaining  untrans- 
formed.  If  that  amount  is  34.2  when  t  =  1  hour  and  11.6  when  t  =  3 
hours,  determine  the  amount  at  the  start,  t  =  0,  and  find  how  many 
hours  will  elapse  before  only  one  per  cent  will  remain. 

270.  Determine  the  shape  of  a  reflector  so  that  all  the  rays  of  light 
coming  from  a  fixed  point  will  be  reflected  in  the  same  direction. 

271.  Find  the  curve  in  which  a  chain  hangs  when  its  ends  are  sup- 
ported at  two  points  and  it  is  allowed  to  hang  under  its  own  weight. 
(See  the  example  solved  in  Art.  57.) 

272.  By  Hooke's  Law  the  amount  an  elastic  string  of  natural  length 
I  stretches  under  a  force  F  is  klF,  k  being  constant.  If  the  string  is  held 
vertical  and  allowed  to  elongate  under  its  own  weight  w,  show  that  the 
elongation  is  \  kwl. 

273.  Assuming  that  the  resistance  of  the  air  produces  a  negative 
acceleration  equal  to  k  times  the  square  of  the  velocity,  show  that  a 
projectile  fired  upward  with  a  velocity  i\  will  return  to  its  starting  point 
with  the  velocity 


—  V, 


gvr 


g  +  kvi2 


g  being  the  acceleration  of  gravity. 

274.    Assuming  that  the  density  of  sea  water  under  a  pressure  of  p 
pounds  per  square  inch  is 

P  =  1  4-  0.000003  p, 


170  Supplementary  Exercises 

show  that  the  surface  of  an  ocean  5  miles  deep  is  about  465  feet  lower 
than  it  would  be  if  water  were  incompressible.  (A  cubic  foot  of  sea 
water  weighs  about  64  pounds.) 

275.  Show  that  when  a  liquid  rotating  with  constant  velocity  is  in 
equilibrium,  its  surface  is  a  paraboloid  of  revolution. 

276.  Find  the  path  described  by  a  particle  moving  in  a  plane,  if  its 
acceleration  is  directed  toward  a  fixed  point  and  is  proportional  to  the 
distance  from  the  point. 


ANSWERS   TO   EXERCISES 


1. 

2. 

3. 

4. 

5. 

6. 
7. 
8. 
9. 
10. 

11. 


12.    - 


Page  5 

^-f^  +  l^  +  C. 

§x*  +  2x*  +  C. 

a  V2^  (2  x  -  3)  +  C. 

x*(fx3  +  |x2  +  fx)+C. 

a*  x  -  2  ax*  +  f  a*  x2  -  f  x*  +  C. 
i  x4  +  I  (a  +  6)  x3  +  |  a&x2  +  C. 
2x  +  31nx  +  C. 
i?/2  +  4?/  +  41n2/  +  C. 
xiCyV^-fx2  -6)  +C. 

In  (x  +  1)  +  C. 
1 


+  C 


x  +  1 

13.  V2x  +  1  +  C. 

14.  |ln(x2  +  2)  +  C. 

15.  V'x?  -  1  +  C. 

16"    "  4  6  (a  +  6x2)2  +  ^ 

17.  -  Ha2  -  *2)§  +  C 

18.  i  In  (a3  +  x3)  +  C. 

19.  |  (x3  -  1)*  +  C. 

20    In  (x2  +  ax  +  b)  +  C. 
v^21.    2  Vx2  +  ax  +  b  +  C. 

22.    -  i-  In  (1  -  at5)  +  C. 
o  <z 


23. 

24. 


25. 


26. 
•27. 

28. 

29. 

30. 


_    1   (a2   _  p)*  +  £ 

x  +  3  In  (x  -  2)  +  C. 
1 


ln(2x2  +  l) 
1 


i(-9 


4  (2  x2  +  1) 

9 


+  c. 


w(n-l)  (xn+a) 
V2  (\/2^  -  V^)11 


^I+C 


11 


+  C 


ir3  _ 


x3-£ln(x3  +  2)  +  C. 

i.T8-fx5  +  ix2  +  C. 


Pages  12,  13 


s 


V  1.    138.  2.    §  0/2 +  30*. 

•^3.    A  =  —  \  gt-  +  100  /  +  60.     It   reaches   the   highest   point   when 
t  =  3.1  sec.,  h  =  215.3  ft. 
4.    sooo  sec. 

171 


172  Answers  to  Exercises 

5.  ^x  =  t2  —  t  +  1,  y  =  t  —  h  P  +  2.  These  are  parametric  equa- 
tions of  the  path.  The  rectangular  equation  is  x2  +  4  xy  +  4  y2  — 
12x-22?/  +  31  =0. 

S6.   About  53  miles.  10.    (-f,  -*£)• 

7.   x  =  ^Vdt2,y  =  ^t2  +  Vot.  11.    6y  =  x*-3x2  +  3rr  +  13. 


4      -->»       4 

y/%.   y  =  2x-\x2-$- 


12.    -12i 
i/'9.   y  =  ex.  *'  14.   About  4  per  cent. 

15.   x  =  Xoekt,  where  x  is  the  number  at  time  t,  Xo  the  number  at  time 
t  =  0,  and  k  is  constant. 

17.  17  minutes.  19.    11.6  years. 

18.  11.4  minutes. 

Pages  18,  19 

^{.    -  (§cos2rr+ssin2rz)  +  C.        ?,  6.    |sin20  +  C. 

/I    5        /2rr-3\    ,    „ 
l/  2.   ^  sin  ( — ^ —  j  +  C.  7.   tan  rr  +  C. 

,/g.    _I  cos(ni+a)  +C.  .        8'    -|cot2x  +  C. 

*4.   3tan^  +  C.  9.    -cscrr  +  C. 

a 

l  h.    —4  csc  7  +  C.  10.    £  sec4  x  +  C. 


11.   2  (csc  |  -  cot  |^  +  C. 


12.  |  sin  (re2  -  1)  +  C. 

13.  |  (tan  3  rr  +  sec  3  re)  +  C. 

14.  tan  re  +  x  —  2  In  (sec  x  +  tan  re)  +  C. 

15.  In  (1  +  sin  x)  +  C. 

16.  0  +  cos2  6  +  C. 

17.  sin  re  +  In  (csc  re  —  cot  x)  +  C. 
^18.  |sin3re  +  C. 

y!9.  itan4rc  +  C. 

j/20.  |tan2x  +  C. 

/21.  -|cos6x  +  C. 

22.  |ln(l  +  2  tan  re)  +  C. 

23.  -§ln(l  -sin 2 re)  +  C. 

24.  -ln(l  +  tan  ax)  +C. 

CL 

^0,       1      •  _,  xV2       r 

"  25.  —j=  sin-1  — —  +  C. 

V2  V3 


Answers  to  Exercises  173 


V26.  -L  tan-  ^  +  C. 

0_     1  xVs 

27.  -  sec"1  — g-  +  C. 

28.  Jtan-12y  +  C. 

„  29.   -i=  In  (x  v7  +  V7  x2  +  l)  +  C. 

V$0.   |sec-ly  +  C. 

Q1         1          2x  +  V3 
-.31.   7=  In t=  +  C. 

4  VS       2  x  -  V  3 

/Z2.hn  (2  x  +  V-i  x-  -  3)  +  C. 
33.    -3  V4t-a?  -  2  sin"1  |  +  C. 

1  34.   2  V^+l  +  3  In  (x  +  Vx2  +  4)  +  C. 

35.    hn(4x2-5)+-^ln2:r~^  +  C- 

5  Vo       2x  +  V5     • 

l/$6.   |  V3x2-9  -  -?=  In  (x  +  Vx7^)  +  C. 
3  V3 


L*. 


37.   sin"1  f ^\  +  C.  46.    -  i  e"**  +  C. 


fc2 


v38.    -V2-sin2x  +  C-  47.  ^- (e2ax  -  e"2") +  2x  +  C. 

39.   tan"1  (sin  x)  +  C.  48.  i  In  (1  +  e3*)  +  C. 

*  40.   sec"1  (tan  x)  +  C.  49.  In  (e*  +  e"x)  +  C. 

41.   In  (sec  x  +  Vsec2  x  +  l) .  _I 

.42.    2Vl-cos.  +  C.  ^5L  4,^^+c. 

43.  -hnf  +  ^  +  C.                          52  ln  LzJL*  +  n 
4       2  -  In  x                                    oz-  ln  i  _|_  ex  +  ^  • 

44.  —  ^  Vcos2x  —  sin2x  +  C.  ,/KO  1    •     .  ,  MN    ,   ^ 

K53.  -  sin  x  eM   +  C. 

1    .   _  x2  a 

45.  -  sin  l  -5  +  C.  54>  tan_!  (e*)  +  c# 

Page  20 

.     1  ,      _,  x  +  3       ~  1    .     ,2a;-  1   .  „ 

L  2 tan  ~^r  +  c-  2-  2 sin  "vT  + 

3.  --^  In  (3  x  +  2  +  V9x2  +  12x  +  6)   +  C. 
v3 

.       1     .       (2x  -  1)V5      „  .  1         _,  (x  -  3)  V6   ,   „ 

4.  —--  sin  » - \-C.  5.-^  sec"1 ^ h  C. 

v5  3  V3  3 


174  Answers  to  Exercises 

6.  1      u,  £±_«  +  c. 
o  —  a       x  +  o 

7.  lln(4a:2_4,-2)+^ln|^i^  +  C. 

8.  %  V3x2-6x  +  l  +  4=M3  (x  -  1)  +  V9x2-18x+3]  +  C. 
3  V3 

9.  ^ln(3x2  +  2x  +  2)  +  — !-=  tan"1  ^t-^  +  C. 
6  i;        3  v  5  v  5 

10.   4=  sec-1  2 *  .t  X  +  5 In  (2x  +  1  +  V4x2  +  4x-l)  +  C. 
V2  V2         * 

01.  — -=L=+c. 

Vx2  -  2  x  +  3 
12.    Vx2  -  x  -  2  +  fin  (2x -  1  +  V4x2  -  4x  -  8)  +  C. 

^3.   -L  In  (4^  +  3~^)  +  C. 
V17       \4ea:  +  3  +  Vl7/ 

Page  25 

1.  —  cosx  +  I  cos3  a;  +  C. 

2.  sin  x  —  |  sin3  x  +  \  sin5  a:  +  C. 

3.  sin  x  —  |  cos3  x  +  f  sin3  x  —  cos  x  +  C. 

4.  —  I  cos3  re  +  -5-  cos5  x  +  C. 

5.  I  sin5  i  x  -  f  sin7 §  a;  +  §  sin9  §  x  +  C. 

6.  TV  sin6  3  0  -  2V  sin8  3  9  +  C. 

7.  -  f  cos3  0  +  cos  0  +  C. 

8.  sin  x  +  \  sin2  x  +  C. 

9.  cos  x  +  In  (esc  £  —  cot  x)  +  C. 
y(0.  *6os2  0  -  i  cos4  0  -  In  cos  0  +  C. 
/ll.  tan  x  +  I  tan3  x  +  C- 

12.  -  (cot  2/  +  f  cot3  y  +  f  cot5  ?/  +  *  cot7  2/  +  £  cot9  2/)  +  C. 

13.  tan  a;  —  x  +  C. 

14.  2tan0  -  sec0-0  +  C. 
05.  fsec^x  +  C. 

^IjB.  x? sec7  2x-\  sec5  2  x  +  \ sec3 2 a;  +  C. 

1/17.  —  ^  esc2  x  —  In  sin  x  +  C. 

*    18.  £  sec6  x  —  f  sec4  x  +  f  sec2  re  +  In  cos  a;  +  C. 

19.  -  £  cot5  x  -  i  cot3  x  +  C. 

20.  §  tan2  a;  +  In  tan  x  +  C. 

*21.   |-  J-  sin  (2 ax)  +  C. 
2       4a 

y22.   |  +  4^  sin  (2  ax)  +  C. 


Answers  to  Exercises  175 


23.  xV  x  —  fa  sin  4  x  —  xV  sin3  2  x  +  C. 

24.  TV  .r  —  3V  sin  2  x  +  2V  sniS  •£  +  C- 

25.  y\  x  —  1  sin  l^  +  ^j sin  4  x  +  ?^  sin3  2x  +  C. 

26.  tan  x  +  sec  x  +  C. 

27.  tan  \  .r  +  C. 

28.  2  f  sin  |  -  cos  |^  +  C. 

29.  I  V^TT^  -  %  In  (x  +  V32  -  a2)  +  C. 

n 

30.  I  Vx2  +  a2  +  I  In  (.r  +  Vx2  +  a2)  +  C. 
A  31.   I  Vx2  +  a-  -  I  In  (3  +  V^  +  fl2)  +  c. 

\S2. /  +  C. 

a2  Vx2  -  a2 

^33.   iln f  +  C. 

a      a  _j_  Va2  _  X2 

V2  r/x  —  r2 

34.  __L^ — *.  +  c. 

ax 

35.  t    l         +  C. 
Va*  -x* 

•^36.    I  (x2  +  a2)1  -  £  (x2  +  o2)f  +  C. 

0  o 

1^7.    -2^±?  +  C. 


a2x 


38. 


^-y^  Vx2  -  4  x  +  5  +  i  In  (x  -  2  +  Vx2  -  4  x  +  5)  +  C. 


39.    n324a?  ^2  -  2x  -  4x2  +  ~  sin-*  i^±_i  +  C. 

Page  30 

1.  tt  +  4x  -  21n(x  -  1)  +  121n(x  -  2)  +  C. 

2.  31nx  -ln(x  +  1)  +  C. 

3.  ln('-»(3!  +  1)+C. 

X 

4.  f +  lnx--^-ln(2x-  1)  --^ln(2x  +  l) +C. 
4  lu  10 

5.  f  In  (x  +  3)  -  £  In  (x  +  1)  -  f  In  (x  +  5)  +  C. 

6.  Hn(2x-  1)  -31n(2x  -3)  +  f  In  (2  x  -  5)  +  C. 

7.  x  +  -  +  In  (X  ~  1)2  +  C 

x  x 


10. 


176  Answers  to  Exercises 

8.   i  In  (_  +  1)  + 1  In  (x  -  1)  -  ^TTf)  +  C- 

Q1    ,     ,n       72x2  +  96x  +  40 
x_81n(:c  +  1) ______ +C. 

11.   ^  +  2x  +  3        n£_-J      ^ 
x3  •  x 

12     i  In  X  ~  2  —  i       g        4-  T 
1J-    2x  +  2       2x2-4+C' 

13'    2  (4  -  x2)  +  C* 

1       x  —  1       1 

14.  x  +  T  In  — — -  —  -  tan-1  x  +  C. 

4      x  +  1      2 

1,  x  +•  1         ,     1    ,       .  2 _ '  —  1       „ 

15.  =  In  -—__—=_  H —  tan-1 — 1-  C. 

*       v  x2  -  x  +  1       V3  V  3 

16.  iln03  +  l)+C. 

17    ______  j    1^    x~  1       2V^.     -i  ______       ^ 

1 '  •    FT T~fT  +  1  ln  TT n-  tan-1  — h  C. 

6  (x  +  1)       4       x  +  1  9  V3 

m  8        i1!  x  —  2  1  x  +  1    .    ~ 

19- i _  +  a  ln     /  =  H 7=  tan-1  — —  +  C. 

x3  -  8      6       Vx2  +  2  x  +  4      2  V3  V3 

20.  3  (x  +  1)*  +  ln  [(x  +  1)*  -  1]  -  V3  tan"1  2  (* +  *_)*  + *  +  C. 

V3 

21.  -i**  +  iaj»-l_*-_A  +  5_i^-t^T  +  C. 

5  4  3  2      x  _  x& 

22.  -^  (ax  +  6)*  -  |^  (ax  +  6)*  +  C. 
o  a  o  a 

23.  2  yx  +  2  -  ln  (x  +  3)  -  2 tan"1  Vx  +  2  +  C. 
v  24.   4  x*  +  2  ln  (x*  -  l)  +  ln  (x*  +  l)  -  2  tan"1  x*  +  C. 

25, 


3.  2  Vx  +  2  -  ln  (x  +  3)  -  2  tan"1  Vx  +  2  +  C. 

4.  4  x*  +  2  ln  (x*  -  l)  +  ln  (x*  +  l)  -  2  tan"1  x* 

5.  i(^  +  l)§  +  H^-l)f  +  C. 


Page  34 


x 


/_.   ^  cos  2  x  +  I  sin  2  x  +  C.  3.   x  sin"1  x  +  Vl  -  x2  +  C. 

/  2.   I  lnx  -  |  +  C.  4.   ^t-i  tan"1-  -  i  x  +  C. 

5.   x  ln  (x  +  Va2  +  x2)  -  Va2  +  x2  +  C. 


Answers  to  Exercises  177 

6.  2  Vx  -  1  In  x  -  4  Vx  -  1  +  4  tan"1  Vx  -  1  +  C. 

t    7.  In  x  In  (In  x)  —  In  x  +  C 

8.  ^  x3  sec"1  x  -  |  Vx*  -  1  -  i  In  (x  +  Vx*  -  1)  +  C. 

9.  x  -  (1  +  e"*)  In  (1  +ex)  +  C. 

•  10.  (x2  -2x  +  2)cx4-C. 

•11.  -  (x3  +  3  x2  +  6  x  +  6)  e"*  +  C. 

ytOt.  x-l.0         2x2-4x  +  l         0       .   ^ 

^fpi2.  — —  sin  2  x cos  2  x  +  C. 

^  13.  |  Vx2  -  a2  -  ^  In  (x  +  Vx2  -  a2)  4-  C. 

yl4.  |  Va2  4-  x2  +  |  In  (x  4-  Va2  +  a?)  +  C. 

Ho.  ^5  (2sin3x-3cos3x)  +  C. 

16.  —  (cos  x  +  sin  x)  +  C. 

17.  -V(sin2x  +  2cos2x)  +  C. 
5 

18.  |  (sec  0  tan  0  +  In  (sec  0  4-  tan  0)]  +  C. 

19.  |  cos  x  —  TV  cos  5  x  +  C. 


1. 

t- 

2. 

2.829. 

•^f. 

1  -*V3 

.    il 

7T 

—   • 

3 

•* 

-20. 

4. 

2. 

5. 

2 

6. 

1.807. 

i  7. 
18. 

0.287^ 

t  9. 

fa. 

10. 

2. 

11. 

00. 

Page  38 

3. 

-  0.630. 

Pages  45,  46 

12. 

7T 

— •    • 

2 

13. 

1 
2/c2 

14. 

0.5493. 

15. 

4^2 

16. 

1.786. 

17. 

0.4055. 

18. 

0.2877. 

19. 

f  d-ln2). 

178 


Answers  to  Exercises 


3. 

A 

t5. 

6. 
7. 
8. 
9. 
11. 
v42. 


7  2. 

4. 


5. 

•«. 

/7. 
8. 


11. 

4  V3(4-  V2). 
|  ^3  +  6. 

9.248. 
7rafr. 
I  a2. 

fV 

9 
2* 


7T 


az 


fa2  V3. 

o2 

\  ^  ~  1). 

a2 
2' 

f  xa2. 
4  a2. 


Pages  49,  50 

13.  2ir  +  *,6ir-f 

14.  4\/3+-136-7r- 

15.  5  ^  -  tan-*  ^ 

17.  37ra2. 

18.  §7ra2. 

19.  tt(62  +  2o6). 

#^20.^  V3. 

21.  37ra2. 

22.  \*ab. 

Page  52 

9.  2a2  (l  +$  V2). 

10.  167r3a2. 

vl3.  |  (t  -  1). 

14.  (10tt  +  9V3)  g 

15.  xa2. 


+  1 


Pages  55,  56 


*4. 


16  _ 

ira2 

~6~' 

7.    f  ?ra3(l  -  cos4  a), 
the  vertical  angle  of  the 
j/&   V-Tra3- 

V9.  tI^«3- 

10.  -3/xa3. 

l/fl.  5  7r2a3. 

^12.  itfsTra3. 


•<f  ('+*-?) 


2. 
3. 
4. 
5. 
6. 


f  a3  tan  a. 
fa3, 
f  7ra62. 
\  ira2h. 
Ah. 


v§. 

1_2 
2  ~  • 

where  a  is  the  radius  of  the  sphere  and  2  a 

cone. 

13. 

f-7ra3. 

14. 

-x33-x2a3. 

15. 

8ttV3. 

16. 

■J  VI. 

Page  59 

8. 

i-o+a  • 

9. 

|  a2/i. 

10. 

4          , 

=,  7ra3. 

3  V2 


Answers  to  Exercises  179 


Page  63 
!  2.   JL(ioVlO-l).  6-   6a; 

*1  In  (2  + VI). 

4.    |  +  Hn2.  |/8.    8a 


*-H 


5.   2.003.  9.   27r2a. 


Page  64 


2^±i  (e  -  1).  i/5.    2  a  [ V2  +  In  (l  +  V2)]. 


3 

a 


4. 

V3* 

|/6.   J^a. 
|/7.    8  a. 
8.    §7ra. 

Pages  66,  67 

3 

.       Va2  -  62\ 

4. 
5. 
6. 

V          Va2  - 
-6^7ra2. 

,    sin  x                   1  • 
o2                  a        / 

7.  -V-7ra2. 

8.  £*-a2  V2(4  -tt). 

9.  87r[V2+ln(l  +  V2)] 
10.   47ra2. 

Page  69 

1.    |7r3a2.  6.  f  7ra3  (1  —  COS  a). 

2-  2  a2.  7<  2a/i(x-2). 

3-  16°2'  _v  8.  §7ra  V7i2  +  4a2. 


9.    ia2A(97r-  16). 


4.  i7ra2(a  +  26  V3). 

5.  fTra3  (2  V2  -  l). 

Pages  71,  72 

•4!   45,000  lbs.  3.   §  uM2. 

*-2.   33,750  lbs.  4.   £  wfc/i2. 

5.  §  wa&2,  where  a  is  the  semi-axis  in  the  surface  and  b  the  vertical 
semi-axi-. 

6.  300,000  w.  7.   40ttw. 

Pages  78-80 

)/\.   \  pa?b,  where  p  is  the  pressure  per  unit  area,  a  the  width,  and  b 
the  height  of  the  door. 

2.  TV  wa*b.  4.   The     intersection     of    the 

3.  -1*2  wbh*  (4  c  +  Sh).  medians. 


180  Answers  to  Exercises 

5.   <fa,0).  12.   y  =  %a. 


(St-—-     Tj  =— •  13.    X  =  |7rV2a. 

°"    x      3tt'   y      3tt 


5- if     »=i^.  13.   x  =  ^V2 

U'  5J 


14.   At  distance   —   from   the 

7T 

bounding  diameter. 


7 

8.    (0,  |f|^)  •  tK    s_«<*  +  4*-l) 


15.   y  = 


4  e  (e2  -  1) 


9.  y-|-  16.    (I a,  fa). 

10.    (&a,  Aa).  17.    (0.399,1.520). 

n.  (|0,^).  m-t- 

19.  On  the  axis  J  of  the  distance  from  the  base  to  the  vertex. 

20.  At  distance  |  a  from  the  plane  face  of  the  hemisphere,  where  a 
is  the  radius. 

21.  (f,0). 

22.  Its  distance  from  the  plane  face  is  T8?  of  the  radius. 

23.  On  the  axis  at  disLance  f  a  (1  +  cos  a)  from    he  vertex,  a  being 
the  radius  and  a  the  angle  of  the  sector. 

24.  (fo,0). 

25.  The  distance  of  the  center  of  gravity  from  the  base  of  the  cylinder 
is  ^  na  tan  a. 

26.  At  the  middle  of  the  radius  perpendicular  to  the  plane  face. 

6V3  +  1 


28.    x  = 


15  V3  -  5 


Pages  82,  83 


2.  2**a*6.  6.    i7ra.3[31n(l  +  V2)  -  V2]. 

3.  ^(l2V3-l).  7.    tira*(3a  +  2sina). 

4.  x  (36^  +  ^6)^.  8-  M' 

5.  \2-ira\ 

10.    §  7ra3  (4  sin  a  —  sin3  a)  tan  a,  where  a  is  the  radius  of  the  cylinder 
and  2  a  the  vertical  angle  of  the  cone. 

Pages  84,  85 

1.  I  a3b,  where  6  is  the  edge  about  which  the  rectangle  is  revolved. 

2.  TV  bh3,  where  6  is  the  base  and  h  the  altitude. 

3.  \  bh3,  where  6  is  the  base  and  h  the  altitude. 

4.  fa4. 

K         6  4    ni 

6.  ^Tra4. 


Answers  to  Exercises  L81 


7.  \  Ma-h,  where  h  is  the  altitude. 

8.  \MaK 

9.  — ^ — ,  where  p  is  the  density. 

10.  I  Ma2,  where  M  is  the  mass  and  a  the  radius. 

12.  Sira4 


v 


1. 


Pages  83,  89 
A:  (6  -  a)* 


2  a 
i^2.   aw  ft.  lbs.,  where  a  is  the  radius  of  the  earth  in  feet. 

v2  -b       a       a 

3.  c  In  7  H 

i\  —  0       Vi       Vi 

4.  25,133  ft.  lbs. 

5.  f  ir/jiPa,  where  a  is  the  radius  of  the  shaft. 

6.  ;= — r  In  - ,  where  /i  is  the  altitude  of  the  cylinder. 
2  irh       a '  J 

7.  -; —  ( r    ,  where  a  and  6  are  the  inner  and  outer  radii. 

47r  \a       b ) 

kh 

8.  — r .  where  a  and  b  are  the  radii  of  the  ends  and  h  the  altitude 
irab 

r 


2.  8.5. 

7.  0.785392. 

8.  1.26. 

9.  4.38. 
10.  21.48. 


•  1.   In  ||. 


2. 

\  na2. 

3. 

¥• 

4. 

7T 

5. 

-1. 

6. 

TTO2 

1  • 

4 

10.  **. 

C 

Pages  95,  96 

11.   4.27. 

12.   0.9045. 

13.   a\  — 

a3X3      a5X5 

3[3  +  5[5  + 

14.    1.91. 

Pages  102, 103 

17         2 

/ .     3. 

15.   4. 

8.   ifa2. 

16.   fa4. 

9.      7T. 

17.    161n2  -^ 

10.    13*. 

18.    fa5. 

11.    3tt. 

19.   fa5. 

12.      IT. 

20.    (f,  -§). 

13.    }. 

21.    (Ya,  -2a) 

14.    §a4. 

■ 

182  Answers  to  Exercises 


1. 

7T04 

■    # 

8 

2. 

7T«2 

~2~' 

3. 

IT 

—  • 

4 

8. 

On  th 

9. 

-V-7ra4 

Pages  107,  108 

4. 


ira3 


6 

6.  £a4(2a  -  sin  2  a). 

7.  A«*(6*  -8). 

2  a  sin  a  .  , . 

— „ irom  the  center. 

11.    a4(4  7r-|f). 
10.    ^vra4.  12.   3  7ra4. 

15.  j3o  Ma2,  M  being  the  mass  and  a  the  radius  of  the  base. 

16.  ia3(3  7r  -4).  18.    T85  7rpa5. 

17.  UaK  l9     (8V2_9)  ^ 


3 


Page  111 

1.  3  Vl4. 

2.  There  are  two  areas  between  the  planes  each  equal  to  2  ma2. 

3.  Two  areas  are  determined  each  equal  to  ira2  v2. 

4.  lira2  V3.  7.   ^7ra2(3  ^3  -  l). 

5.  4.  8.   a2  (tt  -  2). 

6.  8  a2.  9.    8a2tan~4  V2. 

Page  116 

1.  |.  4.   irabc. 

2.  A>  G-   -V-a2/*. 

3.  Its  distance  from  the  base  7.   TV 

is  /a  t1"0- 

Page  121 

1-  vT-  4    Ta^  (2 /i2 -f3 a2). 

2.  f  /i,  where  /*.  is  the  altitude.  60 

3.  7r.  5.   |7ra3.  ♦ 

6.  On  the  axis  of  the  cone  at  the  distance  f  a  (1  +  cos  a)  from  the 
vertex. 

TV 

7.  If  the  two  planes  are  8  =  ±  - ,  the  spherical  coordinates  of  the 

9  7T 

center  of  gravity  are  r  =  —  a,  d  =  0,  $  =  -> 

8.  H^ra5. 


AxsWKlts  TO    Rxercisbs  183 


Page  125 
'   c(c  +1)  a2      [_c       Vc2  +  a'J 


2.^ 


war 

4.  2irkph  (1  —  cos  a),  where  p  is  the  density,  h  the  altitude,  and 
2  a  the  vertical  angle  of  the  cone. 

G.  The  components  along  the  edge  through  the  corner  are  each 
qual  to 

2.1 


a-     L12  1  +  ^J 


Pages  130,  131 

&'  ^da?  +  a;dx      y-°'  b-  *  ^  +  bJ;   dx2  +  *Xdx 

6.   j  cfy  —  y  (x  +  1)  c?x  =  0.  —  4  y  —  0. 

7    ^    ,  0.  9.  ydx  =  x  dy. 


Pages  141-143 


1.  x2  -  f  =  cxy.  lft  2       1 

ID      17   =  CX     —  —  ' 

2.  tan2  x  -  cot2  y  =  c.  x 


3.  f  4-  1  =  c  (x2  -  1). 

4.  *V  +  x2  -  2/2  =  c. 


_3 

11.  ?/  =  cx2e   x, 

12.  a:2?/  =  x  +  q/. 

6.  ^  =  ex2  (/y2  +  1).  14    y=cs-mx_a. 

7.  .r2  +  ?/2  =  ce2ay.  15#    7  X3  =  y  (X7  +  c)i 

8.  xy  =c(y-l).  1  i 

1  16'     3  =  *  +  9+CC      ' 

T  6  -  a  17.   x4  +  4  y  (x2  -  1)*  =  c. 

18.  In  (x2  4-  as,  4-  y2)  4-  -4  tan-i  X-±-^  =  c. 

V3  x  V3 

19.  x2  -  y2  =  ex.  /~  (  25.   y3  =  cex  -  x  -  1. 

20.  ?/2  4-  2  zy  -  c.  \    26.    ey  =  |cx  +  ce~x. 

21.  x4  —  4  xzy  4- 1/4  =  c. 

22.  -0  =  c  -  e"y. 

23.  e*  4-  In  x  =  c. 

24.  j  +  2y  4- In  (x  +  y  -  2)  =  r. 

31.  »  —  le  L   4-  D2   i '  ii  -2    R  sm  at  —  La  ( cos  at  —  e  L>1  J 


27. 

c         X2 

ys=2-2V 

28. 

i/  =  1  x2  4-  c,  or  y  =  ce*. 

29. 

if  =  2  ex  4-  c2. 

30. 

7  =  Ec\l   -  r~to). 

184  Answers  to  Exercises 

32.  y*  =  8  ex~2.  34.   y  =  ex2. 

33.  y2  =  2  ax. 

35.  re  =  a  In f  -f  Va2  -  y2  +  c. 

a  +  v  a2  -  ?/2 

36.  ?/2  +  (x  -  c)2  =  a2. 

37-  y  =  C2el+fce~1' 

38.  r  =  e9. 

39.  r  =  csin0. 

40.  r  =  a  sec  (0  +  c). 

41.  a0  =  Vr2  —  a2  —  a  sec-1  -  +  c. 

a 

42.  y  =  ek. 

43.  A  circle. 

44.  A  straight  line. 

45.  A  circle  with  the  fixed  point  on  its  circumference  or  at  its  center. 

46.  The  logarithmic  spirals  r  =  cekd. 

47.  0.999964. 

Pages  154-156 

1.  y  =  Ci  In  x  —  \  x2  +  c2. 

2.  y  =  x  +  Cixex  -\-  &z. 

3.  y  =  deax  +  ctf-ax. 

4.  y  —  Ci  sin  arc  +  c2  cos  ax. 

5.  t  =  J  \/r> ,.    ,    „  „ds  +  c2. 


c2. 


2  A;  +  Cis 

6.  s  =  4  In  (cie°5'  -  ^~aht)  + 

a2 

7.  y  =  ~  x2  —  —  In  x  +  C2. 

8.  y  =  -L  [ecix+C2  +  e-(W«2)j. 

2  Ci 

9.  i/  =  Ci  +  c^e4*. 

10.  y  =  Cie6X  +  c2e~x. 

11.  y  =  (ci  +  c2x)  e3*. 

12.  y  =  Ci  cos  x  +  c2  sin  x. 

13.  2/  =  Ci  +  c2e-x  +  c3e3x. 

14.  ?/  =  Ciex  -+-  c2e-a:  +  c3  cos  x  +  c4  sin  x. 

15.  y  =  e*  [ci  cos  (x  V2)  +  c2  sin  (x  V2)]. 

i«  -i*r  zV3   .         .    xVjfl 

16.  2/  =  e  'x    Ci  cos  — - (-  c2  sm  — —    • 


Answers  to  Exercises  185 

17.  y  =  c,cx  +  c2e~x  +  c3cx  ^  +  ciC~x  A 

18.  y  =  (ci  +  c-2.c  +  c3x2)  cx. 

19.  y  =  x  +  3  +  Ci  cos  x  +  c2  sin  x. 

20.  */  =  dc-x  +  coc"21  -  £  cx. 

21.  */  =  cie«*  +  cje-**  -  J  x2  -  &  x  -  Tfo. 

22.  y  =  ccx  —  |  (sin  x  +  cos  x). 

23.  */  =  o  +  Cot'21  —  \  x2  +  x. 

24.  y  =  dc"1  +  ear6*  +  £  *  -  A  +  &  e3*. 

25.  j/  =  cicaj:  +  c2c"aI  +  ^-  c". 

26.  2/  =  e*z    ci  cos  — 1-  C2  sin     9        —  —r  (2  sin  2  x  +  3  cos  2  x). 

[x  \/3  x  V3~] 

Ci  cos  — h  c2  sin  — —    —  x3  -+-  x2  —  6. 

28.  */  =  Ciez  +  c2c31  —  ^e2x  sin  x. 

29.  ?/  =  Cie3X  +  c2e~3X  +  /y  e3*  (6  sin  x  —  cos  x). 

30.  #  =  Ci  +  c2x  +  c3x2  +  Cie~x  +  T^g?  (4  cos  4  x  —  sin  4  x). 

31.  1/  =  Ci  cos  2  x  +  (c2  +  \  x)  sin  2  x. 

32.  2/  =  e"*  (ci  +  c2x  +  £  x2)  +  I  cx. 

33.  x  =  Ci  cos  f  +  c2  sin  i  +  2  ifi1  —  c-'), 
y  =  Ci  sin  £  —  c2  cos  £  +  2  (e*  —  e_<)« 

34.  y  =  Ci  cos  i  +  c2  sin  £  —  1, 

x  =  (ci  +  c2)  cos  £  +  (c2  —  Ci)  sin  £  —  3. 

35.  x  =  Cie1  +  c3e~3', 

?/  =  C\e~l  +  3  Cse-3'  +  cos  t. 

36.  x  =  Cie*  +  c2e-<  +  c3  cos  £  +  £4  sin  t, 
y  =  Cie1  +  cie'1  —  c3  cos  I  —  c4  sin  £. 

37.  y  =  x. 

38.  2  ?/*  =  x  +  2. 

39.  »  =  f  *  +  j|  (e-**  -  1). 

40.  s=^ln^ ^ J* 

41.  s  =  b  cos  (ZcO- 

42.  About  7  miles  per  second. 

43.  About  42 £  minutes. 

44.  t  =  V-In  (5  + V24). 

45.  «  =  4=  ln  (9  +  4  v^)« 


TABLE   OF  INTEGRALS 

un  du  =  — —  ,  if  n  is  not  —  1. 
n  +  1 

2.  f^  =  ln«. 

J   u 

r     du  1        _  u 

3.  -t— : — ■  =  -  tan  l  -• 
J  v?  -\-  a?      a  a 

r     du  1    ,     u  —  a 

4.  -= =  x-  In  — — -  • 

J  u2  —  a?      2  a       u  -\-  a 

5.  JV  du  =  eM. 

6.  I  au  du  =  ; 

»/  In  a 

Integrals  of  Trigonometric  Functions 

7.  I  sin  udu  =  —cos  u. 

8.  j  sin2  m  dw  =  |  u  —  \  sin  2  u  =  \  (u  —  sin  u  cos  u). 

9.  J  sin4  it  dit  =  |  u  —  I  sin  2  u  +  sV  sin  4  w. 

10.  f  sin6  it  dit  =  T56  m  —  |  sin  2  it  +  ¥V  sin3  2  it  +  &  sin  4  u. 

11.  I  coswdw  =  sin  w. 

12.  1  cos2  it  du  =  |  w  +  -J  sin  2  it  =  £  (1*  +  sin  it  cos  it). 

13.  J  cos4  it  dit  =  f  w  +  j  sin  2  w  +  g1^  sin  4  -a. 

14.  j  cos6  w  dit  =  T56  u  +  j  sin  2  it  —  ^g-  sin3  2  it  +  /?  sin  4  14. 

15.  I  tanitdit  =  —In  cos  u. 

16.  j  cot  u  du  =  In  sin  u. 

186 


Table  of  Integrals  187 

17.  Caecudu  =  In  (sec  u  +  tan  u)  =  In  tan  [o  +  l)' 

18.  i  Bee1  u  du  =  tan  u. 

19.  f  sec*  u  dti  =  3  sec  u  tan  u  +  h  In  (sec  w  +  tan  it). 

csc  u  du  =  In  (esc  u  —  cot  w)  =  In  tan  ■%• 

21.  |  csc2udu  =  —  cotu. 

22.  f  csc3  udu  =  —  \  csc  w  cot  u  +  5  In  (csc  u  —  cot  u). 

Integrals  containing   Va2  —  u2 

23.  |Vo2-M2dti  =  ^  Va^¥2  +  «   sin-i  H. 
J  2  2a 

24.  f  u-  Va*  -  tf  du  =  3  (2  u2  -  a2)  Va2  -  u2  +  ^  sin"1  -• 
J  8  8  a 

..      r  Va2  -  u2  .  /- 1  ,     a  -  Va2^ 

20.  I   du  =  v  a2  —  w2  +  a  In  — 

»/         u 


u 

26.  I      .  =  sin  x  -• 

^  Va2  -  u2  a 

rt-      r     u*  du  u  ■  /— :    ,   a2    .   _,  u 

27.  I  =  -  3  Va2  -  u2  +  -3  sin"1  -  • 
J  Va2  -  u2  2  2  a 

-  d?i  1  ,     a  —  Va2  —  u2 


28.     f_^===lln 

J  «  Va-  -  u2       a 

29      f '<*"  =  _  X^IZ 

J  u2  Va2  -  u-  a2u 


u 

<hi  Va2  -  u2 


31 


.     f  (a-  -  u2)* du  =  3  (5 a2  -  2 u2)  Va2  -  u»  +  ^  sin'1  -• 
J  o  o  a 

/du  _  u 

(a2  -  u2)§       a2  Va2  -  w2 


Integrals  containing   Vu2  —  a2 

/A 
\A7^~a~2  du  ="  Vu2  -  a2  -  %r  In  (u  +  VM2-a2). 

33.     fu2  VjT^^du  =  |(2u2-a2)  vV  -  a2  -  ^ln  (u  +  Vt^^"2), 


188  Table  of  Integrals 

34.  C^u2  ~a2  du  =  vV  _  a?  _  a  sec-i  Vl . 
J         M  a 

35.  f    ,  ^        du  =  In  (u  +  Vu2  -a2). 
•*  vtt2  -  a2 

36.   r-^^  =  ^V^r72  +  finU  +  V^rr^). 

J  VW2  _  tt2        2  2 

37.     f  — 


dw  1  _,    U 

=  -  sec  x  -' 
u  v  m2  —  a2      a  a 


r         du  Vm2  —  a2 

J  u2  Vu2  -  a2  au 

39.     f  (w2  -  a2)§  dw  =  |  (2  u2 - 5  a2)  vV  -  a2  +  ^ln  (tz+  \V-a2) . 

dw  w 


40 


di 

dfi  — 


(u2  -  a2)*  Vu2  -  a2 


Integrals  containing  Vu2  +  a2 

41.  J  Vu2  +  a2  du  =  '~  Vu2  +  a2  +  |-  In  (u  +  V^2+a2), 

42.  f  w2  VW2  +  a2  rfM  =  |  (2  U2  +  a2)  vV  +  a2  -  |ln  (u  +  Vw2+a2). 

43.  I  ! —  dw  =  v  w2  +  a2  +  a  In ! 

J         u  u 

44.  f    ,  rfM         =ln  (u  +  V^T^"2). 
•J  v  w2  +  a2 

r     u2  du  u     /—t— ; — „       «2  i     /  /— r-; — n\ 

45.  .  =  -  vw2  +  a2  _       in  (w  +  VW2  +  a2)# 

J  Vu2  +  a2       2  2 

._        /*  dw  1  .      V|^2  _|_  a2  _  a 

46.  I  —  =  -  In 

•^   ■?/.  V«!  4-  /i2         Ct 


47 


■  /- 


t  Vw2  _j_  a2      a  u 

du  VV  _f_  a2 


2  Vw2  -|_  a2  a2u 


48.  f  (w2  +  a2)1  du  =  |  (2  w2+5  a2)  VM2  +  a2  +  ^ln  (u+  Vw2+a2). 

49.  J       **  « 


(u2  +  a2)*      a2  Vw2  +  a2 


Table  of  Integrals  189 

Other  Integrals 


50.   /V^S  dx 


ix  +  b 
=  -  V(ax  +  6)  (px  +  q) 


a 


-  bp       °5  In  ( Vp  (ax  +  b)  +  Va  (px  +  q)) 
a  Vap 


1    n r-rr-7 ; — 7       *>p  -  aq  .  V-ap(ax  +  b) 

=  -V(ax  +  b)(px  +  q)  -        . itan  » , — 

a  av—ap  avpx  +  q 

/.    .     ,        eax  (a  sin  bx  —  b  cos  bx) 
cax  sin  bx  ax  = 2   ,   ,  2 ' 

/,     ,        e*"  (b  sin  for  +  a  cos  bx) 
eax  cos  ox  ax  =  2    ,   r ., ' 


190 


Natural  Logarithms 


0-509 


N 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

0 

0.0000 

0.6931 

1.0986 

1.3863 

1.6094 

1.7918 

1.9459 

2.0794 

2.1972 

1 

2.3026 

2.3979 

2.4849 

2.5649 

2.6391 

2.7081 

2.7726 

2.8332 

2.8904 

2.9444 

2 

9957 

3.0445 

3.0910 

3.1355 

3.1781 

3.2189 

3.2581 

3.2958 

3.3322 

3.3673 

3 

3.4012 

4340 

4657 

4965 

5264 

5553 

5835 

6109 

6376 

6636 

4 

6889 

7136 

7377 

7612 

7842 

8067 

8286 

8501 

8712 

8918 

5 

9120 

9318 

9512 

9703 

9890 

4.0073 

4.0254 

4.0431 

4.0604 

4.0775 

6 

4.0943 

4.1109 

4.1271 

4.1431 

4.1589 

1744 

1897 

2047 

2195 

2341 

7 

2485 

2627 

2767 

2905 

3041 

3175 

3307 

3438 

3567 

3694 

8 

3820 

3944 

4067 

4188 

4308 

4427 

4543 

4659 

4773 

4886 

9 

4998 

5109 

5218 

5326 

5433 

5539 

5643 

5747 

5850 

5951 

10 

6052 

6151 

6250 

6347 

6444 

6540 

6634 

6728 

6821 

6913 

11 

7005 

7095 

7185 

7274 

7362 

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7536 

7622 

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7791 

12 

7875 

7958 

8040 

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8520 

8598 

13 

8675 

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8828 

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9053 

9127 

9200 

9273 

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14 

9416 

9488 

9558 

9628 

9698 

9767 

9S36 

9904 

9972 

5.0039 

15 

5.0106 

5.0173 

5.0239 

5.0304 

5.0370 

5.0434 

5.0499 

5.0562 

5.0626 

0689 

16 

0752 

0S14 

0876 

0938 

0999 

1059 

1120 

1180 

1240 

1299 

17 

1358 

1417 

1475 

1533 

1591 

1648 

1705 

1761 

1818 

1874 

18 

1930 

1985 

2040 

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2149 

2204 

2257 

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2364 

2417 

19 

2470 

2523 

2575 

2627 

2679 

2730 

2781 

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29 

2383 

3033 

3083 

3132 

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3230 

3279 

3327 

3375 

3423 

21 

3171 

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3566 

3613 

3660 

3706 

3753 

3799 

3845 

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22 

3936 

3982 

4027 

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4293 

4337 

23 

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4424 

4467 

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25 

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5413 

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5491 

5530 

5568 

26 

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5835 

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5910 

5947 

27 

5984 

6021 

6058 

6095 

6131 

6168 

6204 

6240 

6276 

6312 

28 

6348 

6384 

6419 

6454 

6490 

6525 

6560 

6595 

6630 

6664 

29 

6699 

6733 

6768 

6802 

6S36 

6870 

6904 

6937 

6971 

7004 

30 

7038 

7071 

7104 

7137 

7170 

7203 

7236 

7268 

7301 

7333 

31 

7366 

7398 

7430 

7462 

7494 

7526 

7557 

7589 

7621 

7652 

32 

7683 

7714 

7746 

7777 

7807 

7838 

7869 

7900 

7930 

7961 

33 

7991 

8021 

8051 

8081 

8111 

8141 

8171 

8201 

8230 

8260 

34 

8289 

8319 

8348 

8377 

8406 

8435 

8464 

8493 

8522 

8551 

35 

8579 

8608 

8636 

8665 

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8721 

8749 

8777 

8805 

8833 

36 

8861 

8889 

8916 

8944 

8972 

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9026 

9054 

9081 

9108 

37 

9135 

9162 

9189 

9216 

9243 

9269 

9296 

9322 

9349 

9375 

38 

9402 

9428 

9454 

9480 

9506 

9532 

9558 

9584 

9610 

9636 

39 

9661 

9687 

9713 

9738 

9764 

9789 

9814 

9839 

9865 

9890 

40 

9915 

9940 

9965 

9989 

6.0014 

6.0039 

6.0064 

6.0088 

6.0113 

6.0137 

41 

6.0162 

6.0186 

6.0210 

6.0234 

0259 

0283 

0307 

0331 

0355 

0379 

42 

0403 

0426 

0450 

0474 

0497 

0521 

0544 

0568 

0591 

0615 

43 

0638 

0661 

0684 

0707 

0730 

0753 

0776 

0799 

0822 

0845 

44 

0868 

0890 

0913 

0936 

0958 

0981 

1003 

1026 

1048 

1070 

45 

1092 

1115 

1137 

1159 

1181 

1203 

1225 

1247 

1269 

1291 

46 

1312 

1334 

1356 

1377 

1399 

1420 

1442 

1463 

1485 

1506 

47 

1527 

1549 

1570 

1591 

1612 

1633 

1654 

1675 

1696 

1717 

48 

1738 

1759 

1779 

1800 

1821 

1841 

1862 

1883 

1903 

1924 

49 

1944 

1964 

1985 

2005 

2025 

2046 

2066 

2086 

2106 

2126 

50 

2146 

2166 

2186 

2206 

2226 

2246 

2265 

2285 

2305 

2324 

N 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

e 


N  vnuu,  Logarithms 


101 


500   1009 


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0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

50 

6  2146 

6  2166 

6.21S6 

6  2206 

6.2226 

6.2246 

6.2265 

6.2285 

6.2305 

6  2324 

51 

2344 

2361 

2383 

2403 

2422 

2442 

2461 

2480 

2500 

2519 

52 

2558 

2577 

2596 

2615 

2634 

2672 

2691 

2710 

53 

2729 

2748 

2766 

2  7n."> 

2804 

2823 

28 1 1 

2860 

2879 

2897 

54 

2916 

2934 

2953 

2971 

2989 

3008 

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3014 

3063 

3081 

55 

3099 

3117 

3135 

3154 

3172 

3190 

3208 

3226 

3244 

3261 

56 

3279 

3297 

3315 

3333 

3351 

336N 

3386 

3404 

3421 

3439 

57 

3456 

3474 

3491 

3509 

3526 

35 1 1 

3561 

3578 

3596 

3613 

58 

3630 

3648 

3665 

36S2 

3699 

3716 

3733 

3750 

3767 

3784 

■ 

3S01 

3818 

3835 

3852 

3869 

3886 

3902 

3919 

3936 

3953. 

60 

3969 

30S6 

4003 

4019 

4036 

4052 

4069 

4085 

4102 

4118 

81 

4135 

4151 

4167 

4184 

4200 

4216 

4232 

4249 

4265 

4281 

62 

4297 

4313 

4329 

4345 

4362 

4378 

4394 

4409 

4425 

4441 

63 

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4489 

4505 

4520 

4536 

4552 

4568 

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4615 

4630 

4646 

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4677 

4693 

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4723 

4739 

4754 

65 

4770 

4785 

4800 

4816 

4831 

4846 

4862 

4877 

4892 

4907 

66 

4922 

4938 

4953 

4968 

4983 

4998 

5013 

5028 

5043 

5058 

67 

5073 

50S8 

5103 

5117 

5132 

5147 

5162 

5177 

5191 

5206 

68 

5221 

5236 

5250 

5265 

5280 

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5309 

5323 

5338 

5352 

69 

5367 

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5396 

5410 

5425 

5439 

5453 

5468 

5482 

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70 

5511 

5525 

5539 

5554 

556S 

55S2 

5596 

5610 

5624 

5639 

71 

5653 

5G67 

5681 

5695 

5709 

5723 

5737 

5751 

5765 

5779 

72 

5793 

5806 

5820 

5834 

5848 

5862 

5876 

5889 

5903 

5917 

73 

5930 

5944 

5958 

5971 

5985 

5999 

6012 

6026 

6039 

6053 

74 

6067 

6080 

6093 

6107 

6120 

6134 

6147 

6161 

6174 

6187 

*•  - 

10 

6201 

6214 

6227 

6241 

6254 

6267 

6280 

6294 

6307 

6320 

76 

6333 

6346 

6359 

6373 

63S6 

6399 

6412 

6425 

6438 

6451 

77 

6464 

6477 

6190 

6503 

6516 

6529 

6512 

6554 

6567 

6580 

78 

6593 

6G06 

6619 

6631 

6644 

6657 

0670 

6682 

6695 

6708 

79 

6720 

6733 

6746 

6758 

6771 

67S3 

6796 

6809 

6821 

6834 

80 

6S46 

6859 

6871 

6884 

6896 

6908 

6921 

6933 

6946 

6958 

81 

6970 

69S3 

6995 

7007 

7020 

7032 

7044 

7050 

7069 

7081 

82 

7093 

7105 

7117 

7130 

7142 

7154 

7166 

7178 

7190 

7202 

83 

7214 

7226 

7238 

7250 

7262 

7274 

7286 

7298 

7310 

7322 

84 

7334 

7346 

7358 

7370 

7382 

7393 

7405 

7417 

7429 

7141 

85 

7452 

7461 

7476 

7488 

7499 

7511 

7523 

7534 

7546 

7558 

86 

7569 

7581 

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7604 

7616 

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7639 

7650 

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87 

7685 

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7754 . 

7765 

7776 

7788 

88 

7799 

7811 

7822 

7833 

7845 

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7890 

7901 

89 

7912 

7923 

7935 

7946 

7957 

7968 

7979 

7991 

8002 

8013 

90 

8024 

8035 

8046 

8057 

8068 

8079 

8090 

8101 

8112 

8123 

91 

SI  34 

8145 

8156 

8167 

817S 

8189 

8200 

8211 

8222 

8233 

92 

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8255 

8265 

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8287 

8298 

8309 

8320 

8330 

8341 

93 

8352 

8363 

8373 

8384 

8395 

8405 

8416 

8427 

8437 

8448 

94 

8459 

8469 

8480 

8491 

8501 

8512 

S5_!2 

8533 

8544 

8554 

8565 

8575 

8586 

8596 

8607 

8617 

8628 

8638 

8648 

8659 

96 

B669 

8690 

8701 

8711 

8721 

8732 

8742 

8752 

S763 

97 

B773 

S7S3 

8794 

8804 

8814 

882 1 

8845 

SN.V, 

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98 

8876 

8886 

8896 

8906 

8916 

8926 

8947 

8957 

8967 

99 

8977 

8987 

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9007 

9017 

in  12  7 

9037 

9048 

9058 

168 

100 

907S 

9088 

9098 

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9117 

9127 

9137 

9147 

9157 

9167 

N 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

INDEX 


The  numbers  refer  to  the  pages. 


Approximate  methods,  90-96. 
Area,  by  double  integration,  97. 

bounded  by  a  plane  curve,  47-52. 

derivative  of,  39. 

of  a  surface  of  revolution,  65. 

of  any  surface,  108. 
Attraction,  121. 

Center  of  gravity,  73-80. 
Change  of  variable,  44. 
Constants  of  integration,  1,  128. 
Curves  with  a  given  slope,  7. 
Cylindrical  coordinates,  116. 

Definite  integrals,  36. 

properties  of,  41. 
Differential  equations,  126-156. 

exact,  133. 

homogeneous,  139. 

linear,  136,  147. 

of  the  second  order,  143. 

reducible  to  linear,  137. 

simultaneous,  153. 

solutions  of,  128. 

with  variables  separable,  132. 

Exact  differential  equations,  133. 

Formulas  of  integration,  14. 

Homogeneous    differential    equa- 
tions, 139. 

Infinite  limits,  42. 

Infimte  values  of  the  inteerand,  43. 


Integral,  definite,  36. 

definition  of,  1. 

double,  97. 

indefinite,  36. 

triple,  112. 

Integrals,   containing  ax2  +  bx  -f- 

c,  19. 

v 
containing  (ax  +  b)  fl,  29. 

of  rational  fractions,  26-29. 

of  trigonometric  functions,  21- 

23. 
relation  of  definite  and  indefinite, 

40. 
Integrating  factors,  135 
Integration,  1. 

by  substitution,  15-19. 

constant  of,  1. 

formulas  of,  14. 

geometrical    representation    of, 

36. 
in  series,  94. 
of  rational  fractions,  26-29. 

Length  of  a  curve,  61,  63. 
Limits  of  integration,  36,  42. 
Linear  differential  equations,  136, 
147. 

Mechanical  and  physical  applica- 
tions, 70-89. 
Moment,  72. 

of  inertia,  83. 
Motion  of  a  particle,  5. 


193 


194 


Index 


Order  of  a  differential  equation, 
126. 

Pappus's  theorems,  80. 
Physical  and  mechanical  applica- 
tions, 70-89. 
Pressure,  70. 
Prismoidal  formula,  90. 
Polar  coordinates,  50,  63,  103. 

Rational  fractions,  integration  of, 

26-29. 
Reduction  formulas,  33. 

Separation  of  the  variables,  9,  132. 
Simpson's  rule,  93. 


Spherical  coordinates,  119. 
Summation,  35. 

double,  100. 

triple,  113. 

Trigonometric  functions,  integrals 

of,  21-23. 
Trigonometric  substitutions,  23. 

Variables,  separation  of,  9,  132. 
Volume,  by  double  integration,  99. 

of  a  solid  of  revolution,  52. 

of  a  solid  with  given  area  of  sec- 
tion, 56. 

Work,  85. 


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